(D) The number of moles of water $(n)$ is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.90 \ g}{18 \ g \ mol^{-1}} = 0.05 \ mol$

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(D) The number of moles of water $(n)$ is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.90 \ g}{18 \ g \ mol^{-1}} = 0.05 \ mol$.The temperature $(T)$ is $27 + 273 = 300 \ K$.The pressure $(P)$ in $atm$ is $\frac{32.0 \ Torr}{760 \ Torr \ atm^{-1}} = \frac{32}{760} \ atm$.Using the ideal gas equation $PV = nRT$,we find the volume $(V)$:$V = \frac{nRT}{P} = \frac{0.05 \times 0.082 \times 300}{32 / 760} = \frac{0.05 \times 0.082 \times 300 \times 760}{32} = 29.21 \ L$.The nearest integer is $29$.

Class 11 Chemistry States of Matter Ideal gas equation and Related gas laws

Easy

$A$ box contains $0.90 \ g$ of liquid water in equilibrium with water vapour at $27^{\circ} C$. The equilibrium vapour pressure of water at $27^{\circ} C$ is $32.0 \ Torr$. When the volume of the box is increased,some of the liquid water evaporates to maintain the equilibrium pressure. If all the liquid water evaporates,then the volume of the box must be $.... \ litre$. [nearest integer]
(Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)
(Ignore the volume of the liquid water and assume water vapours behave as an ideal gas.)

JEE MAIN 2022 All JEE MAIN

A
$75$
B
$12$
C
$96$
D
$29$

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