NEET PYQs

Hint

(c) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \longrightarrow 2 \mathrm{NH}_{3}\)
\(1 \mathrm{~Mol}  \mathrm{~NH}_{3}=\frac{3}{2} \mathrm{~mol}  \mathrm{~H}_{2}\)
\(20 \mathrm{~mol}  \mathrm{~NH}_{3}=\frac{3}{2} \times 20 \mathrm{~mol}  \mathrm{~H}_{2}=30 \mathrm{~mol}   \mathrm{~H}_{2}\)
\(\therefore 30\) moles of \(\mathrm{~H}_{2}\) are required.

Hint

(c) Number of atoms
\(=\frac{W}{\text { Molar mass }} \times N_{A} \times\) atomicity
(a) Number of \(\mathrm{Mg}\) atoms \(=\frac{1}{24} \times N_{A} \times 1\)
(b) Number of \(\mathrm{O}\) atoms \(=\frac{1}{32} \times N_{A} \times 2\)
(c) Number of Li atoms \(=\frac{1}{7} \times N_{A} \times 1\)
(d) Number of \(\mathrm{Ag}\) atoms \(=\frac{1}{108} \times N_{A} \times 1\)

Hint

(a)
(1) Mass of water \(=18 \times 1=18 \mathrm{~g}\)
Molecules of water \(=\operatorname{mole} \times N_{A}\) \(=\frac{18}{18} N_{A}=N_{A}\)
(2) Molecules of water \(=\operatorname{mole} \times N_{A}\) \(=\frac{0.18}{18} N_{A}=10^{-2} N_{A}\)
(3) Molecules of water \(=\) mole \(\times N_{A}=10^{-3} N_{A}\)
(4) Moles of water \(=\frac{0.00224}{22.4}=10^{-4}\)

Molecules of water \(=\) mole \(\times N_{A}=10^{-4} N_{A}\)

\(18 \mathrm{~mL}\) of water has maximum number of moles which corresponds to maximum number of molecules.

Hint

(b) Ratio of weight of gases \(=\mathrm{W}_{\mathrm{H}_{2}}: \mathrm{W}_{\mathrm{O}_{2}}=1: 4\)
Ratio of moles of gases \(=n_{\mathrm{H}_{2}}: n_{\mathrm{O}_{2}}=\frac{1}{2}: \frac{4}{32}\)
\(\therefore \quad\) Molar Ratio \(=\frac{1}{2} \times \frac{32}{4}=4: 1\)

Hint

(a) Let atomic weight of element \(X\) is \(x\) and that of element \(Y\) is \(y\).
For \(X Y_{2}, \quad n=\frac{w}{\text { Mol. wt. }}\)
\(0.1=\frac{10}{x+2 y} \Rightarrow x+2 y=\frac{10}{0.1}=100 \dots(i)\)
For \(X_{3} Y_{2}, n=\frac{w}{\text { Mol. Wt. }}\)
\(0.05=\frac{9}{3 x+2 y} \Rightarrow 3 x+2 y=\frac{9}{0.05}=180 \dots(ii)\)
On solving equations (i) and (ii), we get \(y=30\) \(x+2(30)=100 \Rightarrow x=100-60=40\)

Hint

(b) \(16.9 \%\) solution of \(\mathrm{AgNO}_3\) means \(16.9 \mathrm{~g}\) of \(\mathrm{AgNO}_3\) in \(100 \mathrm{~mL}\) of solution.
\(16.9 \mathrm{~g}\) of \(\mathrm{AgNO}_3\) in \(100 \mathrm{~mL}\) solution \(\equiv 8.45 \mathrm{~g}\) of \(\mathrm{AgNO}_3\) in \(50 \mathrm{~mL}\) solution.
Similarly, \(5.8 \%\) of \(\mathrm{NaCl}\) in \(100 \mathrm{~mL}\) solution \(\equiv 2.9 \mathrm{~g}\) of \(\mathrm{NaCl}\) in \(50 \mathrm{~mL}\) solution.
The reaction can be represented as :
\(
\mathrm{AgNO}_3+\mathrm{NaCl} \longrightarrow \mathrm{AgCl}+\mathrm{NaNO}_3
\)
\(
\begin{array}{lll}
\text { Initial } & 8.45 / 170 & 2.9 / 58.5 \\
\text { mole } & =0.049 & =0.049
\end{array}
\)
\(
\begin{array}{lllll}
\text { Final moles } & 0 & 0 & 0.049 & 0.049
\end{array}
\)
\(\therefore\) Mass of \(\mathrm{AgCl}\) precipitated \(=0.049 \times 143.3\) \(=7.02 \approx 7 \mathrm{~g}\)

Hint

(c) \(1.8\) gram of water \(=\frac{6.023 \times 10^{23}}{18} \times 1.8\) \(=6.023 \times 10^{22}\) molecules
18 gram of water \(=6.023 \times 10^{23}\) molecules
18 moles of water \(=18 \times 6.023 \times 10^{23}\) molecules

Hint

(c) According to Avogadro’s hypothesis, the ratio of the volumes of gases will be equal to the ratio of their no. of moles.

No. of moles \(=\frac{\text { Mass }}{\text { Mol. mass }}\)
\(
n_{\mathrm{H}_{2}}=\frac{w}{2} ; n_{\mathrm{O}_{2}}=\frac{w}{32} ; n_{\mathrm{CH}_{4}}=\frac{w}{16}
\)
So, the ratio is \(\frac{w}{2}: \frac{w}{32}: \frac{w}{16}\) or \(16: 1: 2\).

Hint

(a) 1 mole \(\equiv 22.4\) litres at S.T.P.
\(n_{\mathrm{H}_{2}}=\frac{22.4}{22.4}=1 \mathrm{~mol} ; n_{\mathrm{Cl}_{2}}=\frac{11.2}{22.4}=0.5 \mathrm{~mol}\)
Reaction is as,
\(\begin{array}{lcccc} & \mathrm{H}_{2(g)}+\mathrm{Cl}_{2(g)} & \longrightarrow & 2 \mathrm{HCl}_{(g)} \\ \text { Initial } & 1 \mathrm{~mol} & 0.5 \mathrm{~mol} & & 0 \\ \text { Final } & (1-0.5) & (0.5-0.5) & & 2 \times 0.5 \\ & =0.5 \mathrm{~mol} & =0 \mathrm{~mol} & & =1 \mathrm{~mol}\end{array}\)
Here, \(\mathrm{Cl}_{2}\) is limiting reagent. So, 1 mole of \(\mathrm{HCl}_{(g)}\) is formed.

Hint

\(
n_{\mathrm{Mg}}=\frac{1}{24}=0.0416 \text { moles }
\)
\(
n_{\mathrm{O}_2}=\frac{0.56}{32}=0.0175 \mathrm{moles}
\)
The balance equation is:
\(
\mathrm{Mg}+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{MgO}
\)
\(
\text { Initial } \quad 0.0416 \text { moles } \quad 0.0175 \text { moles } \quad 0
\)
\(
\text { Final }(0.0416-2 \times 0.0175) \quad 0 \quad 2 \times 0.0175
\)
\(
=0.0066 \text { moles }\left(\mathrm{O}_2 \text { is limiting reagent. }\right)
\)
\(
\therefore \text { Mass of Mg left in excess }=0.0066 \times 24=0.16 \mathrm{~g}
\)

Hint

(d) Moles of urea \(=\frac{6.02 \times 10^{20}}{6.02 \times 10^{23}}=0.001\) Concentration of solution \(=\frac{0.001}{100} \times 1000=0.01 \mathrm{~M}\)

Hint

(b) Millimoles of solution of chloride
\(
=0.05 \times 10=0.5
\)
Millimoles of \(\mathrm{AgNO}_{3}\) solution \(=10 \times 0.1=1\)
So, the millimoles of \(\mathrm{AgNO}_{3}\) are double than the chloride solution.
\(
\therefore \mathrm{XCl}_{2}+2 \mathrm{AgNO}_{3} \rightarrow 2 \mathrm{AgCl}+X\left(\mathrm{NO}_{3}\right)_{2}
\)

Hint

(c) \(8 \mathrm{~g} \mathrm{H}_{2}\) has 4 moles while the others has 1 mole each.

Moles of \(\mathrm{CO}_2=\frac{44}{44}=1 \quad \mathrm{~N}_{\mathrm{A}} \text {Number of molecules}\)
Moles of \(\mathrm{O}_3=\frac{48}{48}=1 \quad \mathrm{~N}_{\mathrm{A}} \text {Number of molecules}\)
Moles of \(\mathrm{H}_2=\frac{8}{2}=4 \quad 4 \mathrm{~N}_{\mathrm{A}} \text {Number of molecules}\)
Moles of \(\mathrm{SO}_2=\frac{64}{64}=1 \quad \mathrm{~N}_{\mathrm{A}} \text {Number of molecules}\)

Hint

(b)

Given, mole of triatomic gas \(=0.1\)
The number of molecules of triatomic can be calculated using the given formula
Number of molecules \(=\) moles \(\times\) Avogadro number
Since one triatomic gas molecule contains three atoms. So, the total number of atoms can be calculated as
Number of atoms \(=3 \times\) Number of mole \(\times\) Avogadro number
\(
\begin{array}{l}
=3 \times 0.1 \mathrm{~mol} \times 6.022 \times 10^{23} \mathrm{~mol}^{-1} \\
=1.806 \times 10^{23}
\end{array}
\)

Hence, the number of atoms is \(1.806 \times 10^{23}\).

Hint

(b) Given that molar mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}=106 \mathrm{~g}\)
Molarity of solution \(=\frac{25.3 \times 1000}{106 \times 250}\) \(=0.9547 \mathrm{~M}=0.955 \mathrm{~M}\)

\( \mathrm{Na}_{2} \mathrm{CO}_{3} \rightarrow 2 \mathrm{Na}^{+}+\mathrm{CO}_{3}^{2-} \)

\(
\begin{aligned}
&{\left[\mathrm{Na}^{+}\right]=2\left[\mathrm{Na}_{2} \mathrm{CO}_{3}\right]=2 \times 0.955=1.910 \mathrm{~M}} \\
&{\left[\mathrm{CO}_{3}^{2-}\right]=\left[\mathrm{Na}_{2} \mathrm{CO}_{3}\right]=0.955 \mathrm{~M}}
\end{aligned}
\)

Hint

(b) \(\mathrm{H}_{2}+1 / 2 \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}\)
\(\begin{array}{ccc}2 \mathrm{~g} & 16 \mathrm{~g} & 18 \mathrm{~g} \\ 1 \mathrm{~mol} & 0.5 & 1 \mathrm{~mol}\end{array}\)
\(10 \mathrm{~g}\) of \(\mathrm{H}_{2}=5 \mathrm{~mol}\) and \(64 \mathrm{~g}\) of \(\mathrm{O}_{2}=2 \mathrm{~mol}\)
\(\therefore\) In this reaction, oxygen is the limiting reagent so amount of \(\mathrm{H}_{2} \mathrm{O}\) produced depends on that of \(\mathrm{O}_{2}\).
Since \(0.5 \mathrm{~mol}  \text{ of } \mathrm{~O}_{2}\) gives \(1 \mathrm{~mol}  \mathrm{~H}_{2} \mathrm{O}\)
\(\therefore 2  \mathrm{~mol}\) of \(\mathrm{O}_{2}\) will give \(4 \mathrm{~mol}  \mathrm{~H}_{2} \mathrm{O}\)

Hint

\(
\mathrm{C}_3 \mathrm{H}_8+5 \mathrm{O}_2 \longrightarrow 3 \mathrm{CO}_2+4 \mathrm{H}_2 \mathrm{O}
\)
\(
1 \text { vol. } \quad \quad 5 \text { vol. } \quad 3 \text { vol. } \quad 4 \text { vol. }
\)
According to the above equation \(1 \mathrm{~vol}\). or 1 litre of propane requires to \(5 \mathrm{~vol}\). or 5 litre of \(\mathrm{O}_2\) to burn completely.

Hint

\(
\underset{n \text { mol }}{\mathrm{PbO}}+\underset{2 n \text { mol }}{2 \mathrm{HCl}} \rightarrow \underset{n \text { mol }}{\mathrm{PbCl}_2}+{\mathrm{H}_2 \mathrm{O}}
\)
\(
\begin{array}{cc}
\frac{6.5}{224} \mathrm{~mol} & \frac{3.2}{36.5} \mathrm{~mol} \\
=0.029 \mathrm{~mol} & =0.087 \mathrm{~mol}
\end{array}
\)
Formation of moles of lead (II) chloride depends upon the no. of moles of \(\mathrm{PbO}\) which acts as a limiting factor here. So, no. of moles of \(\mathrm{PbCl}_2\) formed will be equal to the no. of moles of \(\mathrm{PbO}\) i.e. 0.029 .

Hint

The percentage composition of Carbon : \(38.71 \%\)
Atomic mass of Carbon: 12
So, Mole ratio \(=\frac{\% \text { composition }}{\text { Atomic mass }}\)
Mole ratio \(=\frac{38.71}{12}=3.22\)
Element Hydrogen:
The percentage composition of Hydrogen : \(9.67 \%\)
Atomic mass of Hydrogen: 1
So, Mole ratio \(=\frac{\% \text { composition }}{\text { Atomic mass }}\)
Mole ratio \(=\frac{9.67}{1}=9.67\)
Element Oxygen
The percentage composition of Oxygen:
\(
100-(38.71+9.67)=51.62 \%
\)

Atomic mass of Oxygen:16
So, Mole ratio \(=\frac{\% \text { composition }}{\text { Atomic mass }}\)
Mole ratio \(=\frac{51.62}{16}=3.22\)
Step 2: Simple ration for \(\mathrm{C}, \mathrm{H}\) and \(\mathrm{O}\).
For carbon (C)
Simple ratio : \(\frac{\text { mole ratio }}{\text { smallest mole ratio }}=\frac{3.22}{3.22}=1\)
Hence, Carbon: 1
For Hydrogen \((\mathrm{H})\)
Simple ratio : \(\frac{\text { mole ratio }}{\text { smallest mole ratio }}=\frac{9.67}{3.22}=3\)
Hence, Hydrogen: 3
For Oxygen (O)
Simple ratio : \(\frac{\text { mole ratio }}{\text { smallest mole ratio }}=\frac{3.22}{3.22}=1\)
Hence, Oxygen: 1
Therefore, the empirical formula of the compound is Option (C): \(\mathrm{CH}_3 \mathrm{O}\)

Hint

(d) Average isotopic mass of \(X\)
\(
\begin{aligned}
&=\frac{200 \times 90+199 \times 8+202 \times 2}{90+8+2} \\
&=\frac{18000+1592+404}{100}=199.96 \text { a.m.u. } \approx 200 \text { a.m.u. }
\end{aligned}
\)

Hint

(a) At STP, \(22.4 \mathrm{~L} \mathrm{H}_{2}=6.023 \times 10^{23}\) molecules
\(
15 \mathrm{~L} \mathrm{H}_{2}=\frac{6.023 \times 10^{23} \times 15}{22.4}=4.033 \times 10^{23}
\)
\(
5 \mathrm{~L} \mathrm{~N}_{2}=\frac{6.023 \times 10^{23} \times 5}{22.4}=1.344 \times 10^{23}
\)
\(
\begin{aligned}
&2 \mathrm{~g} \mathrm{~H}_{2}=6.023 \times 10^{23} \\
&0.5 \mathrm{~g} \mathrm{~H}_{2}=\frac{6.023 \times 10^{23} \times 0.5}{2}=1.505 \times 10^{23} \\
&32 \mathrm{~g} \mathrm{~O}_{2}=6.023 \times 10^{23} \\
&10 \mathrm{~g} \text { of } \mathrm{O}_{2}=\frac{6.023 \times 10^{23} \times 10}{32}=1.882 \times 10^{23}
\end{aligned}
\)

Hint

(b) 1 mole of any element contain \(6.023 \times 10^{23}\) number of molecules.
\(
1 \mathrm{~g} \text { mole of } \mathrm{O}_2=32 \mathrm{~g} \mathrm{~O}_2
\)
\(
16 \mathrm{~g} \text { of } \mathrm{O}_2=0.5 \mathrm{~g} \mathrm{~mole} \mathrm{~O}_2
\)
\(1 \mathrm{~g}\) mole of \(\mathrm{N}_2=28 \mathrm{~g} \mathrm{~N}_2\)
\(7 \mathrm{~g} \mathrm{~N}_2=0.25 \mathrm{~g}\) mole \(\mathrm{N}_2\)
\(1 \mathrm{~g}\) mole of \(\mathrm{~H}_2=2 \mathrm{~g} \mathrm{~H}_2\)
\(2 \mathrm{~g} \mathrm{~H}_2=1.0 \mathrm{~g} \mathrm{~mole} \mathrm{~H}_2\)
\(
1 \mathrm{~g} \mathrm{~mole} \mathrm{~NO}_2=14+16 \times 2=46
\)
\(
16 \mathrm{~g} \text { of } \mathrm{NO}_2=0.35 \mathrm{~mole} \mathrm{~NO}_2
\)
\(
2 \mathrm{~g} \mathrm{~H}_2\left(1 \mathrm{~g} \text { mole } \mathrm{H}_2\right) \text { contain maximum molecules. }
\)

Hint

(a)

Step 1: Given data: We are given a peroxide anhydrous enzyme in which \(\mathrm{Se}\) is present as = \(0.5 \%\) by weight
This means we can say that \(0.5 \mathrm{~g}\) of \(\mathrm{Se}\) is present in \(100 \mathrm{~g}\) of the enzyme.
The atmoic weight of \(\mathrm{Se}\) is given as \(=78.4 \mathrm{~gm}\)

Step 2: Calculation of molecular weight of peroxidase anhydrous enzyme:
\(78.4 \mathrm{~g}\) of \(\mathrm{~Se}\) will be present in,
\(
\begin{array}{l}
=\frac{100}{0.5} \times 78.4 \\
=1.568 \times 10^4 \mathrm{~gm} \text { of peroxidase anhydrous enzyme }
\end{array}
\)

Hence, the minimum molecular weight of peroxidase anhydrous enzyme is calculated as \(1.568 \times 10^4 \mathrm{~gm}\).

Hint

Given:Density of \(\mathrm{HCl}=1.17 \mathrm{~g} / \mathrm{cc}\)

\(
\text { Molarity }=(\text { density/molar mass }) \times 1000 \text { moles per liter }
\)

Molar mass of \(\mathrm{HCl}=[1+35.5]=36.5 \mathrm{~g} / \mathrm{mol}\)

Molarity \(=(1.17 / 36.5) \times 1000=32.05 \mathrm{M}\)
\(\therefore\) Molarity of liquid \(\mathrm{HCl}\) with density equal to \(1.17 \mathrm{~g} / \mathrm{cc}\) is 32.05

Hint

(a) Specific volume (vol. of \(1 \mathrm{~g}\) ) cylindrical virus particle \(=6.02 \times 10^{-2} \mathrm{cc} / \mathrm{g}\)
Radius of virus, \(r=7 Å=7 \times 10^{-8} \mathrm{~cm}\)
Volume of virus \(=\pi r^{2} l\)
\(
=\frac{22}{7} \times\left(7 \times 10^{-8}\right)^{2} \times 10 \times 10^{-8}=154 \times 10^{-23} \mathrm{cc}
\)
wt. of one virus particle \(=\frac{\text { Volume }}{\text { Specific volume }}\)
\(
\Rightarrow \frac{154 \times 10^{-23}}{6.02 \times 10^{-2}} \mathrm{~g}
\)
\(\therefore \quad\) Molecular wt. of virus \(=\mathrm{wt}\). of \(N_{A}\) particle
\(
\begin{aligned}
&=\frac{154 \times 10^{-23}}{6.02 \times 10^{-2}} \times 6.02 \times 10^{-23} \mathrm{~g} / \mathrm{mol} \\
&=15400 \mathrm{~g} / \mathrm{mol}=15.4 \mathrm{~kg} / \mathrm{mol}
\end{aligned}
\)

Hint

(c) \(\mathrm{K}_{3}\left[\mathrm{Cu}(\mathrm{CN})_{2}\right]=3(+1)+x+2(-1)=0\) \(\Rightarrow x=-1\)
As the oxidation no. of ‘ \(\mathrm{Cu}\) ‘ is \(-1\) (-ve), so this complex is unstable and is not formed.

Hint

(b) \(\mathrm{BaCO}_{3} \rightarrow \mathrm{BaO}+\mathrm{CO}_{2}\)
\(197 \cdot 34 \mathrm{~g} \rightarrow 22 \cdot 4 \mathrm{~L}\) at N.T.P.
\(
9 \cdot 85 \mathrm{~g} \rightarrow \frac{22 \cdot 4}{197 \cdot 34} \times 9 \cdot 85=1 \cdot 118 \mathrm{~L}
\)
\(\Rightarrow 9.85 \mathrm{~g}  \mathrm{~BaCO}_{3}\) will produce \(1.118 \mathrm{~L}  \mathrm{~CO}_{2}\) at N.T.P. on the complete decomposition.

Hint

(b)

\(
\mathrm{A}_3\left(\mathrm{BC}_4\right)_2
\)
overall compound should be neutral,Let’s check
Charge on \(\mathrm{A}=3 \times 2=+6\)
Charge on \(\left(\mathrm{BC}_4\right)_2=5+(-8) \times 2=-6\)

Hint

(d) No. of molecules in \(4.25 \mathrm{~g}  \mathrm{~NH}_{3}\) \(=\frac{4.25}{17} \times 6.023 \times 10^{23}=2.5 \times 6.023 \times 10^{22}\)
Number of atoms in \(4.25 \mathrm{~g}  \mathrm{~NH}_{3}=4 \times 2.5 \times 6.023 \times 10^{22}=6.023 \times 10^{23}\)

Hint

(d) Zeros placed left to the number are never significant, therefore the no. of significant figures for the numbers.
\(161 \mathrm{~cm}, 0.161 \mathrm{~cm}\) and \(0.0161 \mathrm{~cm}\) are same, i.e. 3

Note: 161 has three significant figures as all are non-zero digits.
0.161 has three significant figures as zero on the left of the first non-zero digit is not significant.
0.0161 also has three significant figures as zero on the left of the first non zero digit are not significant.

Hint

(a) Quantity of iron in one molecule
\(=\frac{67200}{100} \times 0.334=224.45 \mathrm{~amu}\)
No. of iron atoms in one molecule of haemoglobin \(=\frac{224.45}{56}=4\)

Hint

(a)

\(
4 \mathrm{NH}_{3(g)}+5 \mathrm{O}_{2(g)} \rightarrow 4 \mathrm{NO}_{(g)}+6 \mathrm{H}_2 \mathrm{O}_{(l)}
\)
\(
4 \text { mole }+5 \text { mole } \rightarrow 4 \text { mole }+6 \text { mole }
\)
\(
1 \text { mole of } \mathrm{NH}_3 \text { requires }=\frac{5}{4}=1.25 \text { mole of oxygen }
\)
while 1 mole of \(\mathrm{O}_2\) requires \(=\frac{4}{5}=0.8\) mole of \(\mathrm{NH}_3\).
As there is 1 mole of \(\mathrm{NH}_3\) and 1 mole of \(\mathrm{O}_2\), so all oxygen will be consumed.

Hint

(c) \(\mathrm{Ne}^{2+}(8) \Rightarrow 1 s^{2} 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{1} 2 p_{z}^{1}\)
\(\operatorname{Be}^{+}(3) \Rightarrow 1 s^{2} 2 s^{1}\)
\(\mathrm{Cl}^{-}(18) \Rightarrow 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}\)
\(\operatorname{As}^{+}(32) \Rightarrow 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p_{x}^{1} 4 p_{y}^{1}\)
\(\mathrm{Cl}^{-}\)is not paramagnetic, as it has no unpaired electron.

Hint

(b) Weight of gas \(=0.24 \mathrm{~g}\), Volume of \(\mathrm{gas}=\) \(45 \mathrm{~mL}=0.045\) litre and density of \(\mathrm{H}_{2}=0.089\).
We know that weight of \(45 \mathrm{~mL}\) of \(\mathrm{H}_{2}{=}\)
Density \(\times\) Volume \(=0.089 \times 0.045=4.005 \times 10^{-3} \mathrm{~g}\)
Therefore vapour density
\(
\begin{aligned}
&=\frac{\text { Weight of certain volume of substance }}{\text { Weight of same volume of hydrogen }} \\
&=\frac{0.24}{4.005 \times 10^{-3}}=59.93
\end{aligned}\)

Hint

(c) \(\mathrm{Zn}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{ZnSO}_{4}+\mathrm{H}_{2}\)
\((65 \mathrm{~g}) \quad(22400 \mathrm{~mL})\)
Since \(65 \mathrm{~g}\) of zinc reacts to liberate \(22400 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) at STP, therefore amount of zinc needed to produce \(224 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) at \(\mathrm{STP}\)

\(
=\frac{65}{22400} \times 224=0.65 \mathrm{~g}
\)

Hint

(b) Pressure \(=\frac{\text { Force }}{\text { Area }}\)
Therefore dimensions of pressure \(=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^{2}}\) \(=\mathrm{ML}^{-1} \mathrm{~T}^{-2}\) and dimensions of energy per unit volume

\(
=\frac{\text { Energy }}{\text { Volume }}=\frac{\mathrm{ML}^{2} \mathrm{~T}^{-2}}{\mathrm{~L}^{3}}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}
\)

Hint

(a) Volume of oxygen in 1 Litre of air
\(
\frac{21}{100} \times 1000=210 \mathrm{~mL}
\)
\(\therefore 22400 \mathrm{~mL}\) volume at STP is occupied by oxygen \(=1\) mole
Therefore, number of moles occupied by \(210 \mathrm{~mL}\)
\(
\begin{array}{l}
\frac{210}{22400} \\
=0.0093 \mathrm{~mol}
\end{array}
\)

Hint

(c) Each nitrogen atom has 5 valence electrons, therefore total number of electrons in \(\mathrm{N}_{3}^{-}\) ion is 16. Since the molecular mass of \(\mathrm{N}_{3}\) is 42 , therefore total number of electrons in \(4.2 \mathrm{~g}\) of \(\mathrm{N}_{3}^{-}\)ion
\(
=\frac{4.2}{42} \times 16 \times N_{A}=1.6 N_{A}
\)

Hint

\(
\begin{aligned}
&\text { (a) } 5 \mathrm{~M} \mathrm{~H}_{2} \mathrm{SO}_{4}=10 \mathrm{~N} \mathrm{~H}_{2} \mathrm{SO}_{4} \\
&N_{1} V_{1}=N_{2} V_{2} \Rightarrow 10 \times 1=N_{2} \times 10 \Rightarrow N_{2}=1 \mathrm{~N}
\end{aligned}
\)

Hint

(b) Avogadro’s No., \(N_{A}=6.02 \times 10^{23}\) molecules.
\(6.02 \times 10^{24} \mathrm{~CO}\) molecules \(=10\) moles \(\mathrm{CO}\) \(=10 \mathrm{~g}\) atoms of \(\mathrm{O}=5 \mathrm{~g}\) molecules of \(\mathrm{O}_{2}\)

Hint

(a) Average atomic mass \(=\frac{19 \times 10+81 \times 11}{100}=10.81\)

Hint

(c) If \(1 \mathrm{~L}\) of one gas contains \(N\) molecules, \(2 \mathrm{~L}\) of any gas under the same conditions will contain \(2 \mathrm{~N}\) molecules.

Hint

(d) \(\mathrm{Z}_{2} \mathrm{O}_{3}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{Z}+3 \mathrm{H}_{2} \mathrm{O}\)
Valency of metal in \(\mathrm{Z}_{2} \mathrm{O}_{3}=3\)
\(0.1596 \mathrm{~g}\) of \(\mathrm{Z}_{2} \mathrm{O}_{3}\) react with \(6 \mathrm{~mg}\) of \(\mathrm{H}_{2}\).
\(\left[1 \mathrm{mg}=0.001 \mathrm{~g}=10^{-3} \mathrm{~g}\right]\)
\(\therefore 1 \mathrm{~g}\) of \(\mathrm{H}_{2}\) react with \(=\frac{0.1596}{0.006}=26.6\) g of \(\mathrm{Z}_{2} \mathrm{O}_{3}\)
\(\therefore \quad\) Eq. wt. of \(\mathrm{Z}_{2} \mathrm{O}_{3}=26.6\)
Now, Eq. wt. of \(Z+\) Eq. wt. of \(\mathrm{O}=\) Eq. wt. of \(Z+8=26.6\)
\(\Rightarrow\) Eq. wt. of \(Z=26.6-8=18.6\)
\(\therefore \quad\) At. wt. of \(Z=18.6 \times 3=55.8\)
\(
\left[\mathrm{Eq} . \mathrm{wt}=\frac{\text { Atomic wt. }}{\text { Valency of metal }}\right]
\)

Hint

(a) Here, \(C_{p} / C_{V}=1.4\), which shows that the gas is diatomic.
\(22.4 \mathrm{~L}\) at \(\mathrm{NTP}=6.02 \times 10^{23}\) molecules
\(\therefore \quad 11.2 \mathrm{~L}\) at \(\mathrm{NTP}=3.01 \times 10^{23}\) molecules Since gas is diatomic.
\(\therefore \quad 11.2 \mathrm{~L} \text { at } \mathrm{NTP}=6.02 \times 10^{23} \text { atom }\)

Hint

(c) \(\mathrm{C}_{2} \mathrm{H}_{4}+3 \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\)
\( \quad 28 \mathrm{~g} \quad 96 \mathrm{~g}\)
\(2.8 \mathrm{~kg} \mathrm{~C}_{2} \mathrm{H}_{4}=\frac{96 \mathrm{~g}}{28 \mathrm{~g}} \times 2.8 \mathrm{~kg}\)
\(
=\frac{96}{28} \times 2.8 \times 10^{3} \mathrm{~g}=9.6 \times 10^{3} \mathrm{~g}=9.6 \mathrm{~kg}
\)

Hint

(a) \(1 \mathrm{~mol}\) of \(\mathrm{CO}_{2}=44 \mathrm{~g}\) of \(\mathrm{CO}_{2}\)

\(\therefore \quad 4.4 \mathrm{~g} \mathrm{~CO}_{2}=0.1 \mathrm{~mol} \mathrm{~CO}_{2}=6 \times 10^{22}\) molecules [Since, 1 mole \(\mathrm{~CO}_{2}=6 \times 10^{23}\) molecules]

\(=2 \times 6 \times 10^{22}\) atoms of \(\mathrm{~O}=1.2 \times 10^{23}\) atoms of \(\mathrm{~O}\)

Hint

(a) \(1 \mathrm{~mol}  \mathrm{~CCl}_{4}\) vapour \(=12+4 \times 35.5\) \(=154 \mathrm{~g}=22.4 \mathrm{~L}\)
\(\therefore\) Density of \(\mathrm{CCl}_{4}\) vapour \(=\frac{154}{22.4} \mathrm{~g} \mathrm{~L}^{-1}\)
\(
=6.875 \mathrm{~g} \mathrm{~L}^{-1}
\)

Hint

(d) As we know,
\(22400 \mathrm{~cc}\) of \(\mathrm{N}_{2} \mathrm{O}\) contain \(6.02 \times 10^{23}\) molecules
\(\therefore \quad 1 \mathrm{~cc}\) of \(\mathrm{N}_{2} \mathrm{O}\) contain \(\frac{6.02 \times 10^{23}}{22400}\) molecules
Since in \(\mathrm{N}_{2} \mathrm{O}\) molecule there are 3 atoms
\(\therefore \quad 1 \mathrm{~cc} \mathrm{~N}_{2} \mathrm{O}=\frac{3 \times 6.02 \times 10^{23}}{22400}\) atoms
\(
=\frac{1.8 \times 10^{22}}{224} \text { atoms }
\)
No. of electrons in a molecule of \(\mathrm{N}_{2} \mathrm{O}=7+7+8=22\)
Hence, no. of electrons \(=\frac{6.02 \times 10^{23}}{22400} \times 22\) electrons \(=\frac{1.32}{204} \times 10^{23}\) electrons

Hint

(a) \( \mathrm{Mg}^{2+}+\mathrm{Na}_{2} \mathrm{CO}_{3} \rightarrow \mathrm{MgCO}_{3}+2 \mathrm{Na}^{+}\)
\( \quad 1 \mathrm{~g}\) eq. \( \quad 1 \mathrm{~g}\) eq.
1g eq. of \(\mathrm{Mg}^{2+}=12 \mathrm{~g}\) of \(\mathrm{Mg}^{2+}=12000 \mathrm{~mg}\)
Now, 1000 millieq. of \(\mathrm{Na}_{2} \mathrm{CO}_{3}=12000 \mathrm{~mg}\) of \(\mathrm{Mg}^{2+}\)
\(\therefore \quad 1\) millieq. of \(\mathrm{Na}_{2} \mathrm{CO}_{3}=12 \mathrm{~mg}\) of \(\mathrm{Mg}^{2+}\)

Hint

\(
\begin{array}{|l|l|l|l|}
\hline \text { Element } & \text { Mass percentage } & \text { No. of mole } & \text { Mole ratio } \\
\hline \mathrm{C} & 78 \% & 78 / 12=6.5 & 6.5 / 6.5=1 \\
\hline \mathrm{H} & 22 \% & 22 / 1=22 & 22 / 6.5=3.38=3 \\
\hline
\end{array}
\)

From the calculation, the empirical formula is \(\mathrm{CH_3}\).

Hint

(d) In \({ }_{71}^{175} \mathrm{Lu}\),
Mass number \((A)=175=n+p\)
Atomic number \((Z)=71=p=e^{-}\)
\(\therefore\) Number of protons \(=71\)
Number of neutrons
\(=A-Z=175-71=104\)
Number of electrons \(=71\)

Hint

\(
\begin{aligned}
\mathrm{m} & =\frac{1000 \times \mathrm{M}}{1000 \times \mathrm{d}-\mathrm{MM}_{\mathrm{w}}} \\
\mathrm{m} & =\frac{1000 \times 1}{1000 \times 1.25-1 \times 85} \\
\mathrm{~m} & =\frac{1000}{1165}=0.858
\end{aligned}
\)

Hint

Weight of impure limestone \(=20 \mathrm{~g}\)
Weight of pure limestone \(\left(\mathrm{CaCO}_3\right)=20 \%\) of \(20 \mathrm{~g}\)
\(
\begin{aligned}
& =\frac{20}{100} \times 20 \\
& =4 g \\
& n_{\mathrm{CaCO}_3}=\frac{4}{100}=0.04
\end{aligned}
\)
\(
\underset{n=0.04}{\mathrm{CaCO}_3} \rightarrow \mathrm{CaO}+\underset{n=0.04}{\mathrm{CO}_2}
\)
\(
\begin{aligned}
& n_{\mathrm{CO}_2}=0.04 \\
& W_{C O_2}=0.04 \times 44 \\
& =1.76 \mathrm{~g}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Mass }=\text { Volume } \times \text { Density } \\
& =2.5 \times 2.15 \\
& =5.375 \mathrm{~g}
\end{aligned}
\)

Since 2.5 has two significant figures, so the mass of solution in correct significant figures will be \(5.4 \mathrm{~g}\).

Hint

\(F e_{0.96} \mathrm{O}\)
Let, No. of ions of \(\mathrm{Fe}^{+2}=x\)
No. of ions of \(F e^{+3}=0.96-x\)
Total charge on \(O=-2\)
Now, \((x)(+2)+(0.96-x)(+3)+1(-2)=0\)
\(
\begin{aligned}
& 2 x+2.88-3 x-2=0 \\
& x=0.88
\end{aligned}
\)
\(
\mathrm{Fe}^{2+}=0.88, \mathrm{Fe}^{3+}=0.08
\)
\(\therefore\) Fraction of \(F e^{+3}\) ions \(=\frac{0.08}{0.96}=\frac{1}{12}\).

Hint

Molality is the moles of solute dissolved per \(\mathrm{kg}\) of solvent therefore \(500 \mathrm{~g}\), 1 molal solution contains 0.5 of solute, as
\(
\begin{aligned}
& m=\frac{\text { Moles of solute }}{\text { Mass of solvent }(\text { in } \mathrm{kg})} \\
& 1=\frac{0.5}{\text { Mass of solvent }(\text { in } \mathrm{kg})} \\
& \therefore \text { Mass of solvent (in } \mathrm{kg})=0.5 \\
& =500 \mathrm{~g}
\end{aligned}
\)

Hint

\(
\mathrm{CaCO}_{3(s)}+2 \mathrm{HCl}_{\text {(aq.) }} \rightarrow \mathrm{CaCl}_{2 \text { (aq.) }}+\mathrm{CO}_{2(g)}+\mathrm{H}_2 \mathrm{O}_{(l)}
\)
no. of moles of \(\mathrm{CaCO}_3\) (pure) \(=\frac{1}{2} \times\) mole of \(\mathrm{HCl}\)
\(
\begin{aligned}
& {[\text { Mole }=\text { molarity } \times \text { volume }(\text { in ltr. })]} \\
& =\frac{1}{2} \times 0.5 \times \frac{50}{1000}=0.0125 \\
& \text { weight of } \mathrm{CaCO}_3 \text { (pure) }=\text { mole } \times \text { mol. wt } \\
& =0.0125 \times 100=1.25 \mathrm{~g} \\
& \% \text { purity }=\frac{\text { wt. of pure substance }}{\text { wt. of impure sample }} \times 100 \\
& 95=\frac{1.25}{\text { wt. of impure sample }} \times 100 \\
& \text { wt. of impure sample }=\frac{1.25 \times 100}{95}=1.32 \mathrm{~g}
\end{aligned}
\)

Hint

(a) To find the amount of glucose required to prepare 250 mL of a \(\frac{M}{20}\) aqueous solution, we need to follow these steps:
1. Convert the volume from milliliters to liters, because molarity is expressed in moles per liter \((M)\)
2. Use the equation for molarity:
\(
M=\frac{n}{V}
\)
where:
\(M\) is the molarity (given as \(\frac{1}{20} M\) )
\(n\) is the number of moles of solute
\(V\) is the volume of the solution in liters
3. Rearrange the equation to solve for the number of moles of glucose \((n)\) :
\(n=M \times V\)
4. Convert the moles of glucose to grams using the molar mass of glucose.

Let’s proceed with the calculations:
1. Convert the volume to liters:
\(
250 mL=\frac{250}{1000} L=0.25 L
\)
2. Calculate the moles of glucose:
\(
n=\left(\frac{1}{20}\right) \times 0.25 L=\frac{0.25}{20} mol=0.0125 mol
\)
3. Convert the moles of glucose to grams using the molar mass of glucose:
mass \(=n \times\) molar mass
\(
\text { mass }=0.0125 mol \times 180 g mol ^{-1}=2.25 g
\)
So, the amount of glucose required is: 2.25 g

Hint

(a) To determine which of the given options contains the same number of molecules as 1.0 g of \(H _2\) we need to use the concept of moles and Avogadro’s number. First, let’s calculate the number of moles in 1.0 g of \(H _2\)
The molar mass of \(H _2\) is \(2 g / mol\). Therefore, the number of moles of \(H _2\) in 1.0 g can be calculated as:
Number of moles of \(H _2=\frac{\text { Mass }}{\text { Molar Mass }}=\frac{1.0 g}{2 g / mol }=0.5 mol\)
Next, we need to find out which of the given options corresponds to the same number of moles. Let’s calculate the number of moles for each option:
Option A: 14 g of \(N _2\).
The molar mass of \(N _2\) is \(28 g / mol\). The number of moles of \(N _2\) in 14 g is:
Number of moles of \(N _2=\frac{14 g}{28 g / mol }=0.5 mol\)
Option B: 18 g of \(H _2 O\).
The molar mass of \(H _2 O\) is \(18 g / mol\). The number of moles of \(H _2 O\) in 18 g is:
Number of moles of \(H _2 O =\frac{18 g}{18 g / mol }=1 mol\)
Option C: 16 g of CO.
The molar mass of CO is \(28 g / mol\). The number of moles of CO in 16 g is:
Number of moles of \(CO =\frac{16 g}{28 g / mol }=0.571 mol\)
Option D: 28 g of \(N _2\).
The molar mass of \(N _2\) is \(28 g / mol\). The number of moles of \(N _2\) in 28 g is:
Number of moles of \(N _2=\frac{28 g}{28 g / mol }=1 mol\)
Comparing these calculated values, we can see that 0.5 mol of \(N _2\) (Option A) is equal to 0.5 mol of \(H _2\). Therefore, the correct answer is:
Option A: 14 g of \(N _2\).

Hint

(b) To determine the percentage composition of carbon and hydrogen in the organic compound, we need to calculate the amounts of carbon in \(CO _2\) and hydrogen in \(H _2 O\) produced from the complete combustion of the compound.
First, let’s calculate the amount of carbon in \(CO _2\) :
The molar mass of \(CO _2\) is \(44 g / mol\), and the molar mass of carbon (C) is 12 \(g / mol\). Given that 0.2 g of \(CO _2\) is produced, we can use the following proportion to find the mass of carbon:
\(
\frac{12 g( C )}{44 g\left( CO _2\right)}=\frac{x}{0.2 g\left( CO _2\right)}
\)
Solving for \(x\) (mass of carbon):
\(
x=\frac{12 \times 0.2}{44}=\frac{2.4}{44}=0.0545 g
\)
Next, let’s calculate the amount of hydrogen in \(H _2 O\) :
The molar mass of \(H _2 O\) is \(18 g / mol\), and the molar mass of hydrogen \(( H )\) in a single \(H _2 O\) molecule is \(2 g / mol\). Given that 0.1 g of \(H _2 O\) is produced, we can use the following proportion to find the mass of hydrogen:
\(
\frac{2 g( H )}{18 g\left( H _2 O \right)}=\frac{y}{0.1 g\left( H _2 O \right)}
\)
Solving for \(y\) (mass of hydrogen):
\(
y=\frac{2 \times 0.1}{18}=\frac{0.2}{18}=0.0111 g
\)
Now, we can calculate the percentage composition of carbon and hydrogen in the organic compound, which has a total mass of 0.3 g .
Percentage of carbon:
\(
\% C =\left(\frac{0.0545 g}{0.3 g}\right) \times 100=18.18 \%
\)
Percentage of hydrogen:
\(
\% H =\left(\frac{0.0111 g}{0.3 g}\right) \times 100=3.70 \%
\)
Therefore, the correct answer is:
Option B: \(18.18 \%\) and \(3.70 \%\)

Hint

(b)

\(
\begin{aligned}
& M=\frac{W \times 1000}{M_2 \times V(\text { in } m L)} \\
& W=\frac{M \times M_2 \times V(\text { in } m L)}{1000}=\frac{0.75 \times 36.5 \times 25}{1000} \\
& =0.684 g \text { (Mass of } HCl )
\end{aligned}
\)
\(
\underset{36.5 g}{ HCl }+\underset{40 g}{ NaOH } \rightarrow HCl + NaOH
\)
36.5 g HCl reacts with \(NaOH =40 g\)
0.684 g HCl reacts with \(NaOH =\frac{40}{36.5} \times 0.684 \simeq 0.750 g\)
Amount of NaOH left \(=1 g-0.750 g=0.250 g=250 mg\)

Hint

(a) To determine which option contains the highest number of helium atoms, we need to analyze each option based on the amount of helium it represents and apply Avogadro’s Law as required.
Option A: 4 mol of helium
Using Avogadro’s number, which is approximately \(6.022 \times 10^{23}\) atoms per mole, the number of helium atoms in 4 moles can be calculated as:
\(
4 mol \times 6.022 \times 10^{23} \text { atoms } / mol =24.088 \times 10^{23} \text { atoms }
\)
Option B: 4 u of helium
The atomic mass of helium is approximately 4 u (atomic mass units). Therefore, 4 u represents about 1 mole of helium atoms (since the molar mass of helium is approximately \(4 g / mol\), which equals 4 u ). Thus, this option represents:
\(
1 mol \times 6.022 \times 10^{23} \text { atoms } / mol =6.022 \times 10^{23} \text { atoms }
\)
Option C: 4 g of helium
Similarly, as we’ve established that the molar mass of helium is \(4 g / mol , 4 g\) of helium equates exactly to:
\(
1 mol \times 6.022 \times 10^{23} \text { atoms } / mol =6.022 \times 10^{23} \text { atoms }
\)
Option D: 2.271098 L of helium at STP (Standard Temperature and Pressure)
At STP, one mole of any ideal gas occupies 22.4 L . Therefore, the amount of helium in moles for 2.271098 L can be derived from:
\(
\frac{2.271098 L}{22.4 L / mol } \approx 0.101 mole
\)
Using this to find the number of atoms:
\(
0.101 mol \times 6.022 \times 10^{23} \text { atoms } / mol \approx 6.082 \times 10^{22} \text { atoms }
\)
Conclusion:
Comparing the numbers:
Option A: \(24.088 \times 10^{23}\) atoms
Option B: \(6.022 \times 10^{23}\) atoms
Option C: \(6.022 \times 10^{23}\) atoms
Option D: \(6.082 \times 10^{22}\) atoms
Option A clearly contains the highest number of helium atoms, which is \(24.088 \times 10^{23}\) atoms.

Hint

(b)

\(
\begin{array}{|c|c|c|c|c|}
\hline \text { Element } & \begin{array}{c}
\text { Mass } \\
\text { percentage } \%
\end{array} & \text { No. of moles } & \begin{array}{c}
\text { No. of } \\
\text { moles/Smallest } \\
\text { number }
\end{array} & \begin{array}{c}
\text { Simplest whole } \\
\text { number }
\end{array} \\
\hline \text { A } & 32 \% & \frac{32}{64}=\frac{1}{2} & \frac{1}{2} \times 2 & =1 \\
\hline \text { B } & 20 \% & \frac{20}{40}=\frac{1}{2} & \frac{1}{2} \times 2 & =1 \\
\hline \text { C } & 48 \% & \frac{48}{32}=\frac{3}{2} & \frac{3}{2} \times 2 & =3 \\
\hline
\end{array}
\)
So, empirical formula of
\(
X=\begin{aligned}
& A: B: C \\
& 1: 1: 3
\end{aligned}
\)
\(
\therefore \text { The correct empirical formula of compound X } ABC _3
\)

Hint

(c)

\(
HCOOH \xrightarrow[\text { Dehydrating Agent }]{ H _3 SO _4} CO + H _2 O \binom{ H _2 O \text { abosrbed }}{\text { by } H _2 SO _4}
\)
\(
(\text { moles })_i=\frac{2.3}{46}=\frac{1}{20} \quad \quad \quad 0 \quad \quad \quad 0
\)
\(
\begin{array}{llll}
(\text { moles })_f  & & & &  0 & & & \frac{1}{20} & & \frac{1}{20}
\end{array}
\)
\(
\begin{aligned}
& H _2 C _2 O _4 \xrightarrow{ H _2 SO _4} CO + CO _2+ H _2 O \quad {\left[ H _2 O \text { absorbed by } H _2 SO _4\right]} \\
\end{aligned}
\)
\(
(\text { moles })_i= \frac{4.5}{90}=\frac{1}{20} \quad 0 \quad 0 \quad 0
\)
\(
(\text { moles })_f \quad \quad \quad 0 \quad \quad \quad \frac{1}{20} \quad \frac{1}{20} \quad \frac{1}{20}
\)
\(CO _2\) is absorbed by KOH .
So the remaining product is only CO . moles of CO formed from both reactions
\(
=\frac{1}{20}+\frac{1}{20}=\frac{1}{10}
\)
Left mass of \(CO =\) moles \(\times\) molar mass
\(
=\frac{1}{10} \times 28 =2.8 ~g
\)

Hint

(a)

\(
\begin{aligned}
& \text { Molarity }=\frac{\text { Number of moles of solute }}{\text { Volume of solution (in L) }} \\
& \text { Molality }=\frac{\text { Number of moles of solute }}{\text { Mass of solvent (in kg) }} \\
& \text { Mole fraction }=\frac{\text { Number of moles of component }}{\text { Total number of moles of all components }} \\
& \text { Weight percentage }=\frac{\text { Weight of a component }}{\text { Total weight of solution }} \times 100
\end{aligned}
\)

KEY Concept : Those units which are volume related will be affected by the change in temperature.

The definition of molarity is – Number of moles of solute present in one litre of solution. From definition you can see it is depends on volume which increases with increasing temperature and decreases with decreasing temperature.

All the other three options (Molality, Mole fraction, Weight fraction of solute) are mass related units and temperature has no effect on mass.

Hint

(a) We know mass of \(1 mol\left(6.022 \times 10^{23}\right)\) atoms of carbon \(=12 g\)
If Avogadro number is changed to \(6.022 \times 10^{20}\) then mass of 1 mol of cabon is
\(
=\frac{12 \times 6.022 \times 10^{20}}{6.022 \times 10^{23}}=12 \times 10^{-3} g
\)
\(\therefore\) It would change the mass of one mole of carbon.

Hint

(c) Number of moles of water in \(1000 g=\frac{1000}{18}=55.56 mol\)
Thus, mole fraction of solute \(=\frac{1}{1+55.56}=0.0177\)

Hint

(c)

\(
\begin{aligned}
& 2=\frac{m}{63 \times 0.25} \\
& \Rightarrow m =2 \times 63 \times 0.25=31.5 g
\end{aligned}
\)
Now, if concentrated \(HNO _3\) is \(100 \%\) then it requires 31.5 g . But the original solution of \(HNO _3\) is \(70 \%\) concentrated.
Hence, the mass of \(HNO _3\) required to produce 2.0 M solution
\(
\begin{aligned}
& =\frac{100}{75} \times 31.5 \\
& =45 g \text { of } HNO _3
\end{aligned}
\)

Hint

(c) Normality \(=\frac{98 \times 1.8 \times 10}{49}=36 N\)
\(N _2=0.1 \times 2=0.2 N\)
\(N _2 V_2= N _1 V_1\)
\(\Rightarrow 36 \times V=0.2 \times 1000\)
\(\Rightarrow V=\frac{0.2 \times 1000}{36}=5.55 mL\)

Hint

(b) One molal solution means one mole solute present in \(1 kg(1000 g)\) solvent.
\(\therefore\) mole of solute \(=1\)
Mole of solvent \(\left( H _2 O \right)=\frac{1000}{18}\)
\(
X_{\text {solute }}=\frac{1}{1+\frac{1000}{18}}=0.018
\)

Hint

(c)

\(
\begin{aligned}
&\text { Element \% composition At.Wt Molar Ratio Simplest Ratio }\\
&\begin{array}{lllll}
\text { C } & 40.0 & 12 & 40 / 12=3.33 & 3.33 / 3.33=1 \\
\text { H } & 13.3 & 1 & 13.3 / 1=13.3 & 13.3 / 3 / 33=4 \\
\text { N } & 46.7 & 14 & 46.7 / 14=3.33 & 3.33 / 3.33=1
\end{array}
\end{aligned}
\)
\(
\text { Thus, empirical formula of compound is } CH _4 N \text {. }
\)

Hint

(c)

\(
\begin{aligned}
&\text { From molarity equation }\\
&\begin{aligned}
& M_1 V_1+M_2 V_2=M V \\
& \Rightarrow 1 \times 2.5+0.5 \times 3=M \times 5.5 \\
& \Rightarrow M=\frac{4}{5.5}=0.73 M
\end{aligned}
\end{aligned}
\)