NCERT Exemplar MCQs

Hint

(b) Students A:
Average reading $=\frac{3.01+2.99}{2}=3.0 \mathrm{~g}$
Student B: Average reading $=\frac{3.05+2.95}{2}=3.0 \mathrm{~g}$
For both the students $\mathrm{A}$ and $\mathrm{B}$, average reading is close to the correct reading (i.e., 3.0g). Hence, both recorded by student A are more precise as they differ only by $\pm 0.01$, whereas readings recorded by the student $\mathrm{B}$ are differ by $\pm 0.05$. Thus, the results of student $\mathrm{A}$ are both precise and accurate.
Precision refers to the closeness of various measurements for the same quantity and accuracy is the agreement of a perticular value to the true value of the results. Results of student $\mathrm{A}$ are very close to $3 \mathrm{~g}$.

Hint

(c) The relationship between temperature on Fahrenheit scale and temperature of celsius scale is ${ }^{\circ} \mathrm{C} \times \frac{9}{5}+32=^{\circ} \mathrm{F}$.
Substituting values in the above expression, we get${ }^{\circ} \mathrm{C} \times \frac{9}{5}+32=200^{\circ} \mathrm{F}$
Thus, the temperature is $93.3^{\circ} \mathrm{C}$.
Hence, option $\mathrm{c}$ is correct.

Hint

(c) Molarity, $\mathrm{M}=$ number of moles/volume of solution $\mathrm{M}=\frac{\mathrm{n}}{\mathrm{V}}$
Number of moles, $\mathrm{n}=$ mass/molar mass
Molar mass of $\mathrm{NaCl}=58.5 \mathrm{~g} / \mathrm{mol}$
Mass $=5.85 \mathrm{~g}$
$\mathrm{n}=5.85 / 58.5=0.1 \mathrm{~mol}$
Volume, $\mathrm{V}=500 \mathrm{~mL}=0.5 \mathrm{~L}$
$\therefore \mathrm{M}=0.1 / 0.5=0.2 \mathrm{~mol} / \mathrm{L}$
Hence, option $\mathrm{c}$ is correct.

Hint

(b) If $500 \mathrm{ml}$ of solution is diluted to $1500 \mathrm{ml}$ of solution. The morality can be calculated by the formula $\mathrm{M}_{1} \mathrm{~V}_{1}=\mathrm{M}_{2} \mathrm{~V}_{2}$
Where as $\mathrm{M}_{1}=5 \mathrm{M}, \mathrm{V}_{1}=500, \mathrm{~V}_{2}=1500$ and $\mathrm{M}_{2}=$?
$5 \times 500=1500 \times \mathrm{M}_{2}$
$\mathrm{M}_{2}=2500 / 1500$
$=1.66 \mathrm{M}$
Hence the molarity is $1.66 \mathrm{M}$

Hint

(d) Number of moles $=\mathrm{n}=\frac{\text { weight }}{\text { molecular weight }}$
$1.4 \mathrm{gHe} \Rightarrow \frac{4}{4}=1 \mathrm{~mole}$
2. $46 \mathrm{~g} \mathrm{Na} \Rightarrow \frac{46}{23}=2$ moles
3. $0.4 \mathrm{~g} \mathrm{Ca} \Rightarrow \frac{0.4}{40}=0.01$ moles
4. $12 \mathrm{~g} \mathrm{He} \Rightarrow \frac{12}{4}=3$ moles
Number of moles $\propto$ number of atoms
Therefore $12 \mathrm{gHe}$ contains greatest number of atoms.
Hence, option d is correct.

Hint

(c) Molar mass of glucose $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}=18 \mathrm{~g} / \mathrm{mol}$
concentration of gluocose $=0.9 \mathrm{~g} / \mathrm{L}$
mass of glucose in $1 \mathrm{~L}=0.9 \mathrm{~g}$
Molarity, $\mathrm{M}=\frac{\text { no. of moles }}{\text { volume in } \mathrm{L}}$
no. of moles, $\mathrm{n}=$ mass/molar mass
$\mathrm{n}=\frac{0.9}{180}=5 \times 10^{-3} \mathrm{~mol}$
$\mathrm{M}=\frac{5 \times 10^{-3}}{1}=0.005 \mathrm{M}$
Hence, the correct option is $\mathrm{c}$

Hint

(d) No. of moles of $\mathrm{HCL}=18.25 / 36.5=0.5$
Molality $(\mathrm{m})=\frac{\text { No. of mass of solute }}{\text { Mass of solvent in } \mathrm{kg}}$
Molality $=\frac{0.5 \times 1000}{500}=1 \mathrm{~m}$

Hint

(a) Moles of $\mathrm{H}_{2} \mathrm{SO}_{4}=$ Molarity of $\mathrm{H}_{2} \mathrm{SO}_{4} \times$ Volume of solution (L)
$=0.02 \times 0.1$
$=2 \times 10^{-3}$ moles
No. of $\mathrm{H}_{2} \mathrm{SO}_{4}$ molecules $=2 \times 10^{-3} \times 6.022 \times 10^{23}=12.044 \times 10^{20}$ molecules
Therefore, the correct option is $\mathrm{a}$.

Hint

(b) Molecular formula of carbon dioxide is $\mathrm{CO}_{2}$.
Molar mass of carbon dioxide is $12+2(16)=44 \mathrm{~g} / \mathrm{mol}$.
Mass percentage of carbon in carbon dioxide is $\frac{12}{44} \times 100=27.27 \%$.

Hint

(c) Let’s find the Molar mass of Carbon= 12 .
Molar mass of Hydrogen $=1$.
Molar mass of Oxygen $=16 .$
Mass of $\mathrm{CH}_{2} \mathrm{O}=12+2(1)+16=30$.
Molecular weight of compound given is 180 .
So, the molecular weight is $\frac{180}{30}=6$.
$\Rightarrow$ Molecular formula of compound is $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$.

Hint

(a) Density of solution $=3.12 \mathrm{gmL}^{-1}$ (given)
Volume of solution $=1.5 \mathrm{~mL}$
For a solution,
Mass $=$ volume $\times$ density
$=1.5 m L \times 3.12 g m L^{-1}=4.68 g$
The digit 1.5 has only two significant figures, figures, so the answer must also be limited to two significant figures.
So, it is rounded off to reduce the number of significant figures. Hence, the answer is reported as $4.7 \mathrm{~g}$.

Hint

(c) A compound does not retain the physical or chemical properties of its constituent elements. 

Hint

(a) Total mass of iron and oxygen in reactants $=4(55.85)+3(32)=319.4 \mathrm{~g}$.
Total mass of iron and oxygen in product $=2(2 \times 55.85+3 \times 8)=319.4 \mathrm{~g}$.
The total mass of iron and oxygen in reactants = the total mass of iron and oxygen in the product, therefore, it follows the law of conservation of mass. Hence, option a is correct.

Hint

(b) In this equation,

$\underset{44 \mathrm{~g}}{\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})}+\underset{32 \mathrm{~g}}{\mathrm{O}_{2}(\mathrm{~g})} \rightarrow \underset{44 \mathrm{~g}}{\mathrm{CO}_{2}}(\mathrm{~g})+\underset{18 \mathrm{~g}}{\mathrm{H}_{2} \mathrm{O}}(\mathrm{g})$
i.e., mass of reactants $\neq$ mass of products.
Hence, law of conservation of mass is not followed.

Hint

(b) The element, carbon, combines with oxygen to form two compounds, namely carbon dioxide and carbon monoxide. In $\mathrm{CO}_{2}$, parts by mass of carbon combine with 32 parts by mass of oxygen while in $\mathrm{CO}, 12$ parts by mass of carbon combine with 16 parts by mass of oxygen.

Therefore, the masses of oxygen combine with a fixed mass of carbon (12 parts) in $\mathrm{CO}_{2}$ and $\mathrm{CO}$ are 32 and 16 respectively. These masses of oxygen bear a simple ratio of $32: 16$ or $2: 1$ to each other.
This is an example of law of multiple proportion.

Hint

(a) & (d)

Step 1:
1 mole of $\mathrm{O}_{2}$ gas at STP $=6.022 \times 10^{23}$ molecules of $\mathrm{O}_{2}$ (Avogadro’s number)
1 mole of $\mathrm{O}_{2}$ gas at STP $=32 \mathrm{~g}$ of $\mathrm{O}_{2}$
Step 2:
Molecular weight of oxygen $=2 \times 16=32$
Hence, 1 mole of oxygen gas is equal to the molar weight of oxygen gas as well as the Avogadro’s number.
Hence, options (a) and (d) are correct.

Hint

(b) & (c)
$\mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{NaOH} \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+2 \mathrm{H}_{2} \mathrm{O}$
$0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}=0.1$ mole of $\mathrm{H}_{2} \mathrm{SO}_{4}$
$0.1$ mole $\mathrm{H}_{2} \mathrm{SO}_{4}$ reacts with 2 moles of $\mathrm{NaOH}$
$0.1$ mole of $\mathrm{NaOH}$ will react with $=\frac{0.1}{2}$ moles of $\mathrm{H}_{2} \mathrm{SO}_{4}$
$=0.05$ moles of $\mathrm{H}_{2} \mathrm{SO}_{4}$
Here $\mathrm{NaOH}$ is the limiting reactant
2 moles of $\mathrm{NaOH}$ produce 1 mole of $\mathrm{Na}_{2} \mathrm{SO}_{4}$.
$0.1$ mole of $\mathrm{NaOH}=\frac{0.1}{2}=0.05$ mole
Mass of $\mathrm{Na}_{2} \mathrm{SO}_{4}=0.05 \times\left(\right.$ Molar mass of $\left.\mathrm{Na}_{2} \mathrm{SO}_{4}\right)$
$=0.05 \times(46+32+64)$
$=0.05 \times 142=7.10 \mathrm{~g}$

Volume of solution after mixing $=2 \mathrm{~L}$
Molarity of $\mathrm{Na}_{2} \mathrm{SO}_{4}=\frac{0.05}{2}=0.025 \mathrm{~mol} \mathrm{~L}^{-1}$

Hint

Correct options are (c) and (d)
One mole is the amount of a substance that contains as many particle or entities as there are atoms in exactly $12 \mathrm{~g}$ (or $0.012 \mathrm{~kg}$ ) of the $\mathrm{C}-12$ isotope.
$12 \mathrm{~g} \mathrm{C}=1$ mole of $\mathrm{C}$ and, and $23 \mathrm{~g}=\mathrm{Na}=1$ mole of $\mathrm{Na}$
$28 \mathrm{~g} \mathrm{~N}_{2}=1$ mole of nitrogen $=2 \times 6.022 \times 10^{23}$ atoms of nitrogen
$23 \mathrm{~g} \mathrm{O}_{2}=1$ mole of oxygen $=2 \times 6.022 \times 10^{23}$ atoms of oxygen

Hint

Correct options are (a) and (b)
$20 \mathrm{~g} \mathrm{NaOH}=20 / 40=0.5 \mathrm{~mol} \mathrm{NaOH}$ in $200 \mathrm{ml}$ solution.
Molar concentration of $\mathrm{NaOH}=\frac{20}{40}=\frac{0.500 \mathrm{~mol}}{0.200 \mathrm{~L}}=2.5 \mathrm{M}$
Molar concentration of $\mathrm{KCl}=\frac{0.5 \mathrm{~mol}}{0.200 \mathrm{~L}}=2.5 \mathrm{M}$

Hint

Correct options are (c) and (d)
Number of molecules $\mathrm{N}=\mathrm{n} \times \mathrm{N}_{\mathrm{A}}$, where, $\mathrm{n}=$ number of moles, and $\mathrm{N}_{\mathrm{A}}$ is Avogadro’s number
Number of moles $n=\frac{\text { mass }}{\text { given mass }}=\frac{m}{M}$
So, we can write $\mathrm{N}=\frac{\mathrm{m}}{\mathrm{M}} \times \mathrm{N}_{\mathrm{A}}$
For $16 \mathrm{~g}$ of oxygen, mumber of molecules $\mathrm{N}=\frac{16}{32} \times \mathrm{N}_{\mathrm{A}}=\frac{\mathrm{N}_{\mathrm{A}}}{2}$
(a) For $16 \mathrm{~g}$ of CO, mumber of molecules $\mathrm{N}=\frac{16}{28} \times \mathrm{N}_{\mathrm{A}}=\frac{\mathrm{N}_{\mathrm{A}}}{1.75}$
(b) For 28 g of $\mathrm{N}_{2}$, mumber of molecules $\mathrm{N}=\frac{28}{28} \times \mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{A}}$
(c) For $14 \mathrm{~g}$ of $\mathrm{N}_{2}$, mumber of molecules $\mathrm{N}=\frac{14}{28} \times \mathrm{N}_{\mathrm{A}}=\frac{\mathrm{N}_{\mathrm{A}}}{2}$
(d) For $1 \mathrm{~g}$ of $\mathrm{H}_{2}$, mumber of molecules $\mathrm{N}=\frac{1}{2} \times \mathrm{N}_{\mathrm{A}}=\frac{\mathrm{N}_{\mathrm{A}}}{2}$
So, options $\mathrm{C}$ and $\mathrm{D}$ have same number of molecules as in $16 \mathrm{~g}$ oxygen.

Hint

(c) & (d) are the right options

Mass percent $=\frac{\text { Mass of solute }}{\text { Mass of solution }} \times 100$
Mole Fraction: It is the ratio of number of moles of a particular component to the total number of moles of the solution.

Hint

Correct options are (a) and (d)
The given Dalton’s atomic postulate is as follows:
Compounds are formed when atoms of the different elements combine in a fixed ratio.
This statement is related to the Law of multiple proportion and Law of definite proportion.
Here Avogadro’s law is related to the number of molecules of the different gases whereas the Law of conservation of mass states that the total mass during the reaction is conserved.

Thus, Avogadro’s law and Law of conservation of mass are not related to given Dalton’s atomic theory.

Hint

(a) Correct option is (a)
Mass of 1 mole $=12 \mathrm{gm}$
Mass of $6.022 \times 10^{23}$ atom $=12 \mathrm{gm}$
mass of 1 atom $=\frac{12}{6.023 \times 10^{23}}=1.99 \times 10^{-23} \mathrm{gm}$

Hint

(b) Least precise term $2.5$ or $3.5$ has two significant figures. Hence, the answer should have two significant figures $\frac{2.5 \times 1.25 \times 3.5}{2.01} \approx 5.4415=5.4$

Hint

(a) Symbol for SI Unit of mole is mol.
One mole is defined as the amount of a substance that contains as many particles or entities as there are atoms in exactly $12 \mathrm{~g}(0.012 \mathrm{~kg})$ of the $\mathrm{C}-12$ isotope.

Hint

(b) Molality is the number of moles of solute present in one kilogram of solvent but molarity is the number of moles of solute dissolved in one litre of solution.

Molality is independent of temperature whereas molarity depends on temperature.

Hint

(c)

$\begin{aligned} \text { Mass percent of calcium } &=\frac{3 \times(\text { atomic mass of calcium })}{\text { molecular mass of } \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}} \times 100 \\ &=\frac{120 \mathrm{u}}{310 \mathrm{u}} \times 100=38.71 \% \end{aligned}$

$\text { Mass percent of phosphorus } \begin{aligned} &=\frac{2 \times(\text { atomic mass of phosphorus })}{\text { molecular mass of } \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}} \times 100 \\ &=\frac{2 \times 31 \mathrm{u}}{310 \mathrm{u}} \times 100=20 \% \end{aligned}$

$\begin{aligned} \text { Mass percent of oxygen } &=\frac{8 \times(\text { Atomic mass of oxygen })}{\text { molecular mass of } \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}} \times 100 \\ &=\frac{8 \times 16 \mathrm{u}}{310 \mathrm{u}} \times 100=41.29 \% \end{aligned}$

Hint

(a) According to Gay Lussac’s law of gaseous volumes, gases combine or are produced in a chemical reaction in a simple ratio by volume, provided that all gases are at the same temperature and pressure.

Hint

(a) Yes

(b) According to the law of multiple proportions
(c) $\mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}$
$\quad 2 \mathrm{~g} \quad 16 \mathrm{~g} \quad 18 \mathrm{~g}$
(c) $\mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}$
$\quad 2 \mathrm{~g} \quad 32 \mathrm{~g} \quad 34 \mathrm{~g}$
Here masses of oxygen, (i.e., $16 \mathrm{~g}$ in $\mathrm{H}_{2} \mathrm{O}$ and $32 \mathrm{~g}$ in $\mathrm{H}_{2} \mathrm{O}_{2}$ ) which combine with fixed mass of hydrogen $(2 \mathrm{~g})$ are in the simple ratio i.e., $16: 32$ or $1: 2$

Hint

(a)