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14
MCQ (Multiple Correct Answer)

With regard to the gaseous state of matter which of the following statements are correct ?

A
Complete order of molecules
B
  Complete disorder of molecules
C
Random motion of molecules
D
Fixed position of molecules
15
MCQ (Multiple Correct Answer)

Which of the following figures does not represent 1 mole of dioxygen gas at STP?

A
16 g of gas
B
22.7 L of gas
C
$6.022 \times 10^{23}$ dioxygen molecules
D
11.2 L of gas
16
MCQ (Multiple Correct Answer)

Under which of the following conditions applied together, a gas deviates most from the ideal behaviour?

A
Low pressure
B
High pressure
C
Low temperature
D
High temperature
17
MCQ (Multiple Correct Answer)

Which of the following changes decrease the vapour pressure of water kept in a sealed vessel ?

A
Decreasing the quantity of water
B
Adding salt to water
C
Decreasing the volume of the vessel to one-half
D
Decreasing the temperature of water
18
Subjective

If 1 g of each of the following gases are taken at STP, which of the gases will occupy (a) greatest volume and (b) smallest volume?

$$\mathrm{CO}, \mathrm{H}_2 \mathrm{O}, \mathrm{CH}_4, \mathrm{NO}$$

Explanation

From Avogadro's law, we know that

Volume of 1 mole of the gas = graw molecular mass $=22.4 \mathrm{~L}$ at STP

Volume occupied by $28 \mathrm{~g} \mathrm{CO}(1 \mathrm{~mol} \mathrm{CO})=22.4 \mathrm{~L}$ at STP

$\left(\because\right.$ Molar mass of $\left.\mathrm{CO}=12+16=28 \mathrm{~g} \mathrm{~mol}^{-1}\right)$

$\therefore \quad$ Volume occupied by $1 \mathrm{~g} \mathrm{~CO}=\frac{22.4}{28} \mathrm{~L}$ at STP

Similarly, volume occupied by $1 \mathrm{~g} \mathrm{~H}_2 \mathrm{O}=\frac{22.4}{18} \mathrm{~L}$ at STP

$\left(\because\right.$ Molar mass of $\left.\mathrm{H}_2 \mathrm{O}=(2 \times 1)+16=18 \mathrm{~g} \mathrm{~mol}^{-1}\right)$

Volume occupied by $1 \mathrm{~g} \mathrm{CH}_4=\frac{22.4}{16} \mathrm{~L}$ at STP

$\left(\because\right.$ Molar mass of $\left.\mathrm{CH}_4=12+(4 \times 1)=16 \mathrm{~g} \mathrm{~mol}^{-1}\right)$

Volume occupied by $1 \mathrm{~g} \mathrm{NO}=\frac{22.4}{30} \mathrm{~L}$ at STP

$\left(\because\right.$ Molar mass of $\left.\mathrm{NO}=14+16=30 \mathrm{~g} \mathrm{~mol}^{-1}\right)$

Thus, $1 \mathrm{~g} \mathrm{CH}_4$ will occupy maximum volume while 1 g of NO will occupy minimum volume at STP.