Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. 20 dm ^ 3 of SO _ 2 diffuses through the porous partition in 60 seconds. The volume of O _ 2 in dm ^ 3 which diffuses under the similar

Correct answer: C. 14 . 1 As per Graham's law, V _ SO _ 2 / t _ SO _ 2 V _ O _ 2 / t _ O _ 2 = p _ SO _ 2 p _ O _ 2 M _ O _ 2 M _ SO _ 2 = M _ O _ 2 M _ SO _ 2 2 0 dm ^ 3 / 6 0 sec V _ O _ 2 / 3 0 sec = (3 2)/(6 4) = (1)/(1 · 4 1 4) 2 0 × 3 0 6 0 × V _ O _ 2 = (1)/(1 · 4 1 4) ⇒ V _ O _ 2 = 1 0 × 1 · 4 1 4 = 14 . 14 dm ^ 3

JEE Main 201419 AprilOnline

Exam: JEE Main
Year: 2014
Date: 19 April
Mode: Online
Subject: Chemistry
Chapter: States of Matter

JEE Main 2014 · 19 April · Online · Chemistry · States of Matter

Sulphur dioxide and oxygen were allowed to diffuse through a porous...

Sulphur dioxide and oxygen were allowed to diffuse through a porous partition.

20{dm}^{3}

{SO}_{2}

diffuses through the porous partition in

60

seconds. The volume of

{O}_{2}

{dm}^{3}

which diffuses under the similar condition in

30

seconds will be (atomic mass of sulphur

= 32u

);

Options

A $28 . 2$
B $10 . 0$
C $14 . 1$
D $7 . 09$

Correct answer

C. $14 . 1$

Step-by-step solution

As per Graham's law,

\dfrac{{\text{V}}_{{\text{SO}}_{2}} / {\text{t}}_{{\text{SO}}_{2}}}{{\text{V}}_{{\text{O}}_{2}} / {t}_{{\text{O}}_{2}}} = \dfrac{{\text{p}}_{{\text{SO}}_{2}}}{{\text{p}}_{{\text{O}}_{2}}} \sqrt{\dfrac{{\text{M}}_{{\text{O}}_{2}}}{{\text{M}}_{{\text{SO}}_{2}}}} = \sqrt{\dfrac{{\text{M}}_{{\text{O}}_{2}}}{{\text{M}}_{{\text{SO}}_{2}}}}

\dfrac{2 0 {\text{dm}}^{3} / 6 0 \text{sec}}{{\text{V}}_{{\text{O}}_{2}} / 3 0 \text{sec}} = \sqrt{\dfrac{3 2}{6 4}} = \dfrac{1}{1 \cdot 4 1 4}

\dfrac{2 0 \times 3 0}{6 0 \times {\text{V}}_{{\text{O}}_{2}}} = \dfrac{1}{1 \cdot 4 1 4} ⇒ {\text{V}}_{{\text{O}}_{2}} = 1 0 \times 1 \cdot 4 1 4

= 14 . 14{dm}^{3}

Practice States of Matter on Quantrex Academy →

More from States of Matter

Two cylinders, both fitted with frictionless pistons, are filled with mixtures of He and… JEE Advanced 2026 Molar volume (V_m) of a van der Waals gas can be calculated by expressing the van der… JEE Advanced 2025 At the sea level, the dry air mass percentage composition is given as nitrogen gas… JEE Main 2025 For a reaction, N_2 O_ 5(~g) arrow 2 NO_ 2(~g) +(1)/(2) O_ 2(~g) in a constant volume… JEE Main 2025 A closed vessel contains 10 ~g of an ideal gas X at 300 ~K , which exerts 2 atm pressure.… JEE Advanced 2024 A gas has a compressibility factor of 0 . 5 and a molar volume of 0 . 4 dm ^ 3 mol ^ - 1… JEE Advanced 2023 A certain quantity of real gas occupies a volume of 0 . 15 dm ^ 3 at 100atm and 500K when… JEE Main 2023 At 600K , the root mean square ( rms ) speed of gas X (molar mass = 40 ) is equal to the… JEE Main 2023 Full States of Matter list All JEE Main PYQs