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← JEE Advanced 2014 · Paper 2
ChemistrySingle correct+3 / -1Question 35 of 60

JEE Advanced 2014 · Paper 2, Chemistry question 35

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X and Y are two volatile liquids with molar weights of 10 g mol⁻¹ and 40 g mol⁻¹ respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.

A horizontal tube of length L = 24 cm with cotton wool soaked in X plugging the left end and cotton wool soaked in Y plugging the right end; a dashed ring marks the initial formation of the product at distance d from the X end.

The value of d in cm (shown in the figure), as estimated from Graham's law, is

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Answer and worked solution

Official answer: C

Graham's law. Rate of diffusion ∝1/M\propto1/\sqrt M:
rXrY=4010=2.\frac{r_X}{r_Y}=\sqrt{\frac{40}{10}}=2.

Meeting point. In the same time X travels twice as far as Y:
d=22+1×24=16 cm.d=\frac{2}{2+1}\times24=16\text{ cm}.

(C)