Practice Corner MCQs

Hint

(c) To calculate the number of neutrons, we can subtract the mass number of the atom from the atomic number of that atom. We also know that atomic number is represented as Z and atomic mass as A , neutrons are represented as N and protons are represented as P . Oxygen possesses atomic number 8 , mass number or atomic mass 17 . So, using the formula to calculate the number of neutrons in one atom of oxygen:
\(
N=A-Z \Rightarrow N=17-8 \Rightarrow N=9
\)
Now, to calculate the total number of neutrons in \(12 \times 10^{25}\) atoms of oxygen, we can simply multiply the number of neutrons in one atom with this, doing the same we get:
Total number of neutrons:
\(
\Rightarrow n=9 \times 12 \times 10^{25} \Rightarrow n=108 \times 10^{25}
\)
We are also given with Avogadro’s number in the question, so in terms of Avogadro’s number:
\(
\Rightarrow n=\frac{108 \times 10^{25}}{6 \times 10^{23}} \Rightarrow n=1800 N_A
\)

Hint

(a) As we know that the number of nucleons comprising an atom is found out by using mass of one nucleon.
Given that the mass of one atom is \(3.32 \times 10^{-23} g\).
We know that the nucleon can be either a proton or neutron.
We know that the mass of one nucleon is approximately given as: \(1.67 \times 10^{-24} \mathrm{~g}\).
Now we have the mass of one atom and the mass of one neutron so in order to find the number of nucleons in one atom let us divide the weight of an atom by the mass of the nucleon.
So, no of nucleons in one such given atom is:
\(
=\frac{\text { Weight of one atom }}{\text { Weight of one nucleon }}
\)
Let us substitute the values to find the no of atoms.
\(
=\frac{3.32 \times 10^{-23} g}{1.67 \times 10^{-24} g}=1.98 \times 10=19.8 \cong 20
\)
So, the number of nucleons in one atom is around 20
Therefore, number of nucleons in two atoms is:
\(
=20 \times 2=40
\)
Hence, the number of nucleons (neutrons and protons) present in 2 atoms of the element is 40.

Hint

(b) No. of electrons present in \(\mathrm{O}=8\) and \(\mathrm{P}=15\)
Electron present in 1 molecule \(=15+4 \times 8+3=50\)
Electron present in 1 mole \(=50 \mathrm{~N}_{\mathrm{A}}\)
\(
\text { Total mole }=\frac{9.5}{(31+16 \times 4)}=\frac{9.5}{95}
\)
Total no. of electrons in \(\mathrm{PO}_4^{3-}=50 \times \frac{9.5}{95} \mathrm{~N}_{\mathrm{A}}=5 \mathrm{~N}_{\mathrm{A}}\)

Hint

(d)

\(
\begin{aligned}
& \text { Molar mass of } \mathrm{HNO}_3=1+14+16 \times 3=63 \mathrm{~amu} \\
& \text { no. of moles }=\frac{\text { given mass }}{\text { molar mass }}=\frac{126}{63}=2 \\
& 1 \text { mole of } \mathrm{HNO}_3=3 \text { atoms of oxygen } \\
& 2 \text { mole of } \mathrm{HNO}_3=2 \times 3=6 \text { atoms of oxygen } \\
& 1 \text { mole }=6.022 \times 10^{23} \text { or } \mathrm{N}_{\mathrm{A}} . \\
& \text { no of moles of oxygen }=\frac{\text { given atoms }}{\text { Avagadro no. }\left(\mathrm{N}_{\mathrm{A}}\right)}=\frac{6}{\mathrm{~N}_{\mathrm{A}}}
\end{aligned}
\)

Hint

(c) We know that the atomic number of Sulphur is 16 and its atomic mass is 32 amu.
Now, we have to find out how many Sulphur ions are present in 96 amu .
For this first we will calculate the number of ions as:
Number of ions = given mass/atomic mass= \(\frac{96 a m u}{32 a m u}=3\)
Here, we can see that the ion is represented as \(S^{2-}\). That means each sulphide ion has 2 negative charges present in this ion.
As we know that 1 electron has charge \(=1.602 \times 10^{-19} \mathrm{C}\)
Total charge of the sulphide ion represented here \(=\)
\(
\begin{array}{rl}
3 \times 2 & \times 1.062 \times 10^{-19} \\
& =9.6 \times 10^{-19} \mathrm{C}
\end{array}
\)

Hint

(d) First of let’s discuss sodium and ethanol. Sodium is alkali metal which belongs to the s-block element and has the atomic number as 11 and the mass number as 23 . It is very soft and so soft that it could be easily cut with a knife and in its outermost shell it has one electron and has a strong tendency to lose the electron to acquire the stable electronic configuration of the nearest noble gas i.e. the neon.
Sodium is alkaline-earth metal which belongs to the s-block element and has the atomic number as 20 and the mass number as 40 . In its outermost shell it has two electrons and has a strong tendency to lose the electron to acquire the stable electronic configuration of the nearest noble gas i.e. the neon.
Now considering the statement as;
We have been given that the number of the calcium atoms is the same as that of the sodium atoms. So, first we will find out the no of atoms i.e. the moles of sodium.
Mass of sodium \(=46 \mathrm{~g}\) (given)
Molecular mass of sodium \(=23\)
Then, the no. of moles of sodium is as;
\(
\begin{aligned}
& \text { mole }=\frac{\text { given mass }}{\text { molecular mass }} \\
& \Rightarrow \frac{46}{23} \\
& \Rightarrow 2
\end{aligned}
\)
So, therefore, the no of sodium atoms present is 2 .
Now, the mass of the calcium atoms can be calculated as;
mass of calcium atoms \(=2 \times 40\) (molecular mass of \(\mathrm{Ca}=40)\)
\(\Longrightarrow 80 \mathrm{~g}\)

Hint

(c) No. of moles of \(\mathrm{H}_2 \mathrm{O}(l)=\frac{54}{18}\)
\(
\left(d=1.0 \mathrm{~g} / \mathrm{mL} \text { for } \mathrm{H}_2 \mathrm{O}\right)
\)
\(H\) has no neutron
\(
\therefore \text { no. of neutrons in } \mathrm{H}_2 \mathrm{O}=3 \times 8 \times N_A
\)

Hint

(c) The formula used:
\(
\text { No. of moles }=\frac{\text { Mass of the Substance in grams }}{\text { Molar mass of a Substance }}=\frac{\text { Number of Atoms or Molecules }}{6.022 \times 10^{23}}
\)
Complete step by step answer:
The atomic number of magnesium is 12 which means that in a non-ionized state the number of electrons in magnesium is \(12 . \ln \mathrm{Mg}^{2+}\) the number of electrons will be 10.

Calculating the number of moles of \(48 \mathrm{~g} \mathrm{Mg} g^{2+}\) will be \(\frac{48}{24}=2\).
Now for calculating the no. of electrons, we multiply the no. of moles with Avogadro’s number \(=2 \times \mathrm{N}_{\mathrm{A}}\)
The total number of electrons in 48 g of \(M g^{2+}\) will \(=2 \times \mathrm{N}_{\mathrm{A}} \times 10=20 \mathrm{~N}_{\mathrm{A}}\)

Hint

(b) weight of deutron is 2 gm so \(\mathrm{mol}=5 / 2=2.5\)
no of neutron is \(2.5 N_A\)

Hint

(b) Calculate the molar mass of Cisplatin and then divide by Avogadro’s number to find the mass of one molecule.
Step 1: Calculate the molar mass of \(\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\)
Molar mass of \(\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\) is the sum of the atomic masses of each atom in the molecule.
\(
\begin{aligned}
& \text { Molar mass }=195+2(14+3(1))+2(35.5) \\
& \text { Molar mass }=195+2(17)+71 \\
& \text { Molar mass }=195+34+71 \\
& \text { Molar mass }=300 \frac{\mathrm{~g}}{\mathrm{~mol}}
\end{aligned}
\)
Step 2: Calculate the mass of one molecule
Divide the molar mass by Avogadro’s number.
\(
\begin{aligned}
& \text { Mass of one molecule }=\frac{300 \frac{\mathrm{~g}}{\mathrm{~mol}}}{6.022 \times 10^{23} \frac{\text { molecules }}{\mathrm{mol}}} \\
& \text { Mass of one molecule }=4.98 \times 10^{-22} \frac{\mathrm{~g}}{\text { molecule }}
\end{aligned}
\)
Solution: The mass of one molecule of Cisplatin is \(4.98 \times 10^{-22} \mathrm{~g}\).

Hint

(d) Moles of \(\mathrm{C}_9 \mathrm{H}_8 \mathrm{O}_4=\frac{360 \times 10^{-3}}{180}=2 \times 10^{-3}\)
Number of O -atoms \(=2 \times 10^{-3} \times \mathrm{N}_{\mathrm{A}} \times 4\)
\(
=8 \times 6.02 \times 10^{20}=4.81 \times 10^{21}
\)

Hint

(b) \(\text { Moles }=\frac{\text { Volume (at STP) }}{22.4}\)
\(
\text { Also, moles }=\frac{\text { Weight }}{\text { Mol. Mass }}
\)
Thus, \(\frac{\text { Volume }(\text { at STP })}{22.4}=\frac{\text { Weight }}{\text { Mol. Mass }}\)
\(
\frac{5.6}{22.4}=\frac{20}{x}
\)
Hence, \(x=80\)

Hint

(b) Use the ratio of hydrogen atoms to oxygen atoms in the chemical formula to find the number of moles of oxygen atoms.

Step 1: Find the mole ratio between hydrogen and oxygen atoms in \(\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4\)
In one molecule of \(\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4\), there are 12 hydrogen atoms and 4 oxygen atoms.
The ratio of hydrogen to oxygen atoms is \(\frac{12}{4}=\frac{3}{1}\).
For every 3 moles of hydrogen atoms, there is 1 mole of oxygen atoms.

Step 2: Calculate the moles of oxygen atoms
Given 6 moles of hydrogen atoms.
Moles of oxygen atoms \(=\frac{1}{3} \times\) moles of hydrogen atoms.
Moles of oxygen atoms \(=\frac{1}{3} \times 6=2\) moles.
Solution: The number of moles of oxygen atoms in the sample is 2 moles.

Hint

(a) Use the ideal gas law to find the number of moles of \(O_3\), then multiply by 3 to find the number of moles of oxygen atoms.

Step 1: Calculate the number of moles of \(\mathrm{O}_3\)

Use the ideal gas law: \(P V=n R T\)
Solve for \(n: n=\frac{P V}{R T}\)
Substitute the given values:
\(n=\frac{8.21 \mathrm{~atm} \cdot 3 \mathrm{~L}}{0.0821 \frac{\mathrm{Latm}}{\mathrm{mol} \cdot \mathrm{K}} \cdot 300 \mathrm{~K}}\)
\(n=\frac{24.63}{24.63} \mathrm{~mol}\)
\(n=1 \mathrm{~mol}\)

Step 2: Calculate the number of moles of oxygen atoms
Each mole of \(\mathrm{O}_3\) contains 3 moles of oxygen atoms.
Multiply the number of moles of \(O_3\) by 3 :
\(n_O=3 \cdot n_{O_3}\)
\(n_O=3 \cdot 1 \mathrm{~mol}\)
\(n_O=3 \mathrm{~mol}\)

Solution: The total number of moles of oxygen atoms is 3 mol .

Hint

(d) From Avogadro’s law
\(6.023 \times 10^{23}\) atoms or Avogadro, s number of an element contains mass \(=\) Atomic mass of the element
\(\because 3.011 \times 10^{23}\) atoms of an element weight 1.15 g
\(\therefore 6.023 \times 10^{23}\) atoms of an element will weight
\(
=\frac{1.15 \times 6.023 \times 10^{23}}{3.011 \times 10^{23}}=23
\)
Hence, atomic mass of the element is 23 .

Hint

\(
\begin{aligned}
&\text { (d) Atomic weight of an element }\\
&\begin{aligned}
& x=6.643 \times 10^{-23} \times N_A=40 \\
& \text { no. of moles of } x=\frac{20 \times 1000}{40}=500
\end{aligned}
\end{aligned}
\)

Hint

(d) Divide the total mass by the mass of one atom.

Step 1: Calculate the number of atoms
Divide the total mass by the mass of one atom:
Number of atoms \(=\frac{\text { Total mass }}{\text { Mass of one atom }}\)
Number of atoms \(=\frac{1 \mathrm{~g}}{3.9854 \times 10^{-23} \mathrm{~g}}\)
Number of atoms \(=2.509 \times 10^{22}\)

Solution: The number of atoms in 1 g of element A is \(2.509 \times 10^{22}\).

Hint

(d) Calculate the moles of hydrogen atoms in each option and compare them.
Step 1: Calculate moles of H in 5.0 moles of \(\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_4\)
Moles of \(\mathrm{H}=5.0 \times 2=10.0 \mathrm{moles}\)

Step 2: Calculate moles of H in 1.1 moles of \(\mathrm{C}_3 \mathrm{H}_8 \mathrm{O}_3\)
Moles of \(\mathrm{H}=1.1 \times 8=8.8\) moles

Step 3: Calculate moles of H in 1.5 moles of \(\mathrm{C}_6 \mathrm{H}_8 \mathrm{O}_6\)
Moles of \(\mathrm{H}=1.5 \times 8=12.0\) moles

Step 4: Calculate moles of H in 4.0 moles of \(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2\)
Moles of \(\mathrm{H}=4.0 \times 4=16.0\) moles

Step 5: Compare the moles of H
\(
16.0>12.0>10.0>8.8
\)
Solution: The largest mass of hydrogen atoms is in 4.0 moles of \(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2\).

Hint

(d) Calculate the number of oxygen atoms in each sample and compare them.
Step 1: Calculate the mass of \(\mathrm{H}_2 \mathrm{O}\).
Mass of \(\mathrm{H}_2 \mathrm{O}\) is:
\(10 \mathrm{~mL} \times 1 \frac{\mathrm{~g}}{\mathrm{~mL}}=10 \mathrm{~g}\)

Step 2: Calculate the moles of \(\mathrm{H}_2 \mathrm{O}\).
Moles of \(\mathrm{H}_2 \mathrm{O}\) are:
\(\frac{10 \mathrm{~g}}{18 \frac{\mathrm{~g}}{\mathrm{~mol}}}=\frac{5}{9} \mathrm{~mol}\)

Step 3: Calculate the number of oxygen atoms in \(\mathrm{H}_2 \mathrm{O}\).
Number of oxygen atoms in \(\mathrm{H}_2 \mathrm{O}\) is:
\(\frac{5}{9} \mathrm{~mol} \times 1 \frac{\text { atom O}}{\text { molecule } \mathrm{H}_2 \mathrm{O}} \times 6.022 \times 10^{23} \frac{\text { molecules }}{\mathrm{mol}}=3.346 \times 10^{23}\) atoms
Step 4: Calculate the number of oxygen atoms in \(\mathrm{V}_2 \mathrm{O}_5\).
Number of oxygen atoms in \(\mathrm{V}_2 \mathrm{O}_5\) is:
\(0.1 \mathrm{~mol} \times 5 \frac{\text { atoms } \mathrm{O}}{\text { molecule } \mathrm{V}_2 \mathrm{O}_5} \times 6.022 \times 10^{23} \frac{\text { molecules }}{\mathrm{mol}}=3.011 \times 10^{23}\) atoms

Step 5: Calculate the moles of \(\mathrm{O}_3\).
Moles of \(\mathrm{O}_3\) are:
\(\frac{12 \mathrm{~g}}{48 \frac{\mathrm{~g}}{\mathrm{~mol}}}=0.25 \mathrm{~mol}\)

Step 6: Calculate the number of oxygen atoms in \(\mathrm{O}_3\).
Number of oxygen atoms in \(\mathrm{O}_3\) is:
\(0.25 \mathrm{~mol} \times 3 \frac{\text { atoms } \mathrm{O}}{\text { molecule } \mathrm{O}_3} \times 6.022 \times 10^{23} \frac{\text { molecules }}{\mathrm{mol}}=4.5165 \times 10^{23}\) atoms
Step 7: Calculate the number of oxygen atoms in \(\mathrm{CO}_2\).
Number of oxygen atoms in \(\mathrm{CO}_2\) is:
\(12.044 \times 10^{22}\) molecules \(\times 2 \frac{\text { atoms } \mathrm{O}}{\text { molecule } \mathrm{CO}_2}=2.4088 \times 10^{23}\) atoms

Step 8: Compare the number of oxygen atoms.
Comparing the number of oxygen atoms:
\(\mathrm{H}_2 \mathrm{O}: 3.346 \times 10^{23}\)
\(\mathrm{V}_2 \mathrm{O}_5: 3.011 \times 10^{23}\)
\(\mathrm{O}_3: 4.5165 \times 10^{23}\)
\(\mathrm{CO}_2: 2.4088 \times 10^{23}\)

Solution: The minimum number of oxygen atoms is in \(12.044 \times 10^{22}\) molecules of \(\mathrm{CO}_2\).

Hint

 (a) (I) 0.5 mole \(\mathrm{O}_3=24 \mathrm{~g} \mathrm{O}_3\)
(II) 0.5 g atom of oxygen \(=8 \mathrm{~g}\)
(III) \(\frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} \times 32=16 \mathrm{~g} \mathrm{O}_2\)
(IV) \(\frac{5.6}{22.4} \times 44 \mathrm{~g} \mathrm{CO}_2=11 \mathrm{~g} \mathrm{CO}_2\)

Hint

(a)

\(
\begin{aligned}
& \text { Mass of a drop of water }=0.0018 \mathrm{~g} \\
& \text { Moles of water }=\frac{0.0018}{18}=10^{-4} \text { mole } \\
& \text { Number of water molecules in one drop }=10^{-4} \mathrm{~N}_{\mathrm{A}} \\
& \text { Number of water molecules in two drops }=2 \times 10^{-4} \times 6.023 \times 10^{23}=12.046 \times 10^{19}
\end{aligned}
\)

Hint

(c) No. of protons in \({ }_6 \mathrm{C}^{14}=6\);

No. of neutrons in \({ }_6 \mathrm{C}^{14}=8\);
As per given new atomic mass of

\(
{ }_6 \mathrm{C}^{14}=12+4=16
\)

(As the mass of \(e^{-}\)is negligible as compared to neutron and proton)
\(\%\) increase in mass \(=\frac{16-14}{14} \times 100=14.28\)

Hint

(b)
\(
\text { Mass of } \mathrm{NaCl}=25 \times \frac{8.775}{100}=2.193 \mathrm{~g}
\)
\(
\begin{aligned}
& \text { Number of } \mathrm{NaCl} \text { units }=\frac{2.193}{58.5} \times \mathrm{N}_{\mathrm{A}} \\
& =2.25 \times {10}^{22}
\end{aligned}
\)

Hint

(b) Step 1: Calculate the number of moles of sucrose
The number of moles is given by the mass divided by the molar mass:
\(n=\frac{m}{M}\)
\(n=\frac{25.6 \mathrm{~g}}{342.3 \mathrm{~g} / \mathrm{mol}}\)
\(n=0.0748 \mathrm{~mol}\)

Step 2: Calculate the number of molecules of sucrose
The number of molecules is given by the number of moles multiplied by Avogadro’s number:
\(N=n \times N_A\)
\(N=0.0748 \mathrm{~mol} \times 6.022 \times 10^{23}\) molecules \(/ \mathrm{mol}\)
\(N=4.50 \times 10^{22}\) molecules
Step 3: Calculate the number of hydrogen atoms

Each molecule of sucrose contains 22 hydrogen atoms:
\(N_H=N \times 22\)
\(N_H=4.50 \times 10^{22}\) molecules \(\times 22\) atoms / molecule
\(N_H=9.91 \times 10^{23}\) atoms

Solution: The number of hydrogen atoms present in 25.6 g of sucrose is \(9.91 \times 10^{23}\).

Hint

(a) Formula used: The formula used can be written as:
\(
\text { Number of atoms of any element }=\frac{\text { Weight of element in molecule }}{\text { Atomic mass of the element }}
\)
Complete step by step answer:
The given quantities can be written as:
Molecular weight of caffeine \(=194 g\)
Mass percent of nitrogen \(=28.9 \%\)
Atomic weight of nitrogen \(=14 \mathrm{~g}\)
Thus, mass of nitrogen element in one mole of caffeine
\(
=28.9 \% \text { of } 194 g=\frac{28.9}{100} \times 194 g=56.06 g
\)
Number of nitrogen atoms in caffeine can be calculated by dividing the mass of nitrogen in a molecule of caffeine with the atomic mass of nitrogen.
Number of nitrogen atoms in one molecule of caffeine \(=\frac{56.06}{14}=4\).