Some Basic Concepts of Chemistry's Question 38 | Chemistry XI - Part 1 | NCERT Exemplar Questions

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MCQ (Single Correct Answer)

Assertion (A) The empirical mass of ethene is half of its molecular mass.

Reason (R) The empirical formula represents the simplest whole number ratio of various atoms present in a compound.

Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

$A$ is true but $R$ is false.

$A$ is false but $R$ is true.

Both A and R are false.

MCQ (Single Correct Answer)

Assertion (A) One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom.

Reason (R) Carbon-12 isotope is the most abundant isotope of carbon and has been chosen as standard.

Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

Both $A$ and $R$ true but $R$ is not the correct explanation of $A$.

$A$ is true but $R$ is false.

Both A and R are false.

MCQ (Single Correct Answer)

Assertion (A) Significant figures for 0.200 is 3 where as for 200 it is 1. Reason (R) Zero at the end or right of a number are significant provided they are not on the right side of the decimal point.

Both $A$ and $R$ are true and $R$ is correct explanation of $A$.

Both $A$ and $R$ are true but $R$ is not the correct explanation of $A$.

$A$ is true but $R$ is false.

Both A and R are false.

MCQ (Single Correct Answer)

Assertion (A) Combustion of 16 g of methane gives 18 g of water. Reason (R) In the combustion of methane, water is one of the products.

Both $A$ and $R$ are true but $R$ is not the correct explanation of $A$.

$A$ is true but $R$ is false.

$A$ is false but $R$ is true.

Both A and R are false.

Subjective

A vessel contains 1.6 g of dioxygen at STP ( $273.15 \mathrm{~K}, 1 \mathrm{~atm}$ pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate

(a) volume of the new vessel.

(b) number of molecules of dioxygen.

Explanation

$$\text{(a)} \quad p_1=1 \mathrm{~atm}, p_2=\frac{1}{2}=0.5 \mathrm{~atm}, T_1=273.15, V_2=?, V_1=?$$

$$\begin{aligned} & 32 \mathrm{~g} \text { dioxygen occupies }=22.4 \mathrm{~L} \text { volume at STP } \\ & \therefore 1.6 \mathrm{~g} \text { dioxygen will occupy }=\frac{22.4 \mathrm{~L} \times 1.6 \mathrm{~g}}{32 \mathrm{~g}}=1.12 \mathrm{~L} \end{aligned}$$

$$V_1=1.12 \mathrm{~L}$$

$$\begin{aligned} &\text { From Boyle's law (as temperature is constant), }\\ &\begin{aligned} p_1 V_1 & =p_2 V_2 \\ V_2 & =\frac{p_1 V_1}{p_2} \end{aligned} \end{aligned}$$

$$=\frac{1 \mathrm{~atm} \times 1.12 \mathrm{~L}}{0.5 \mathrm{~atm}}=2.24 \mathrm{~L}$$

$$\begin{aligned} \text { (b) } \text { Number of moles of dioxygen } & =\frac{\text { Mass of dioxygen }}{\text { Molar mass of dioxygen }} \\ n_{\mathrm{O}_2} & =\frac{1.6}{32}=0.05 \mathrm{~mol} \end{aligned}$$

$$\begin{aligned} 1 \mathrm{~mol} \text { of dioxygen contains } & =6.022 \times 10^{23} \text { molecules of dioxygen } \\ \therefore \quad 0.05 \mathrm{~mol} \text { of dioxygen } & =6.022 \times 10^{23} \times 0.05 \text { molecule of } \mathrm{O}_2 \\ & =0.3011 \times 10^{23} \text { molecules } \\ & =3.011 \times 10^{22} \text { molecules } \end{aligned}$$

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