JEE Main 2014 · 19 Apr · Shift 1 · Chemistry Q37

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ChemistrySingle correct+4 / -1Question 37 of 90

Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. 20 dm $^3$ of $\ce{SO2}$ diffuses through the porous partition in 60 seconds. The volume of $\ce{O2}$ in dm $^3$ which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur = 32 u) :

7.09

14.1

10.0

28.2

Answer and worked solution

Official answer: B

Graham's law: $\dfrac{r_{\ce{O2}}}{r_{\ce{SO2}}}=\sqrt{\dfrac{64}{32}}=\sqrt2$ .

$\ce{SO2}$ rate: $\frac{20}{60}$ dm³/s. In 30 s the $\ce{O2}$ volume is
$\sqrt2\times\frac{20}{60}\times30=14.1\text{ dm}^3.$

Topics:Kinetic theory of gases

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JEE Main 2014 · 19 Apr · Shift 1, Chemistry question 37