States of Matter - Result Question 51
Passage
$X$ and $Y$ are two volatile liquids with molar weights of $10 \mathrm{~g} \mathrm{~mol}^{-1}$ and 40 g $\mathrm{mol}^{-1}$ 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 $L=24 \mathrm{~cm}$, as shown in the figure. The tube is filled with an inert gas at 1 atm pressure and a temperature of $300 K $. Vapours of $X$ and $Y$ react to form a product which is first observed at a distance $d \mathrm{~cm}$ 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.
(2014 Adv.)
51. The value of $d$ in $cm$ (shown in the figure), as estimated from Graham’s law, is
(a) 8
(b) 12
(c) 16
(d) 20
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Answer:
Correct Answer: 51. (c)
Solution:
This problem can be solved by using the concept of Graham’s law of diffusion according to which rate of diffusion of non-reactive gases under similar conditions of temperature and pressure are inversely proportional to square root of their density.
$ \text { Rate of diffusion } \propto \dfrac{1}{\sqrt{\text { molar weight of gas }}} $
Let distance covered by $X$ is $d$, then distance covered by $Y$ is $24-d$.
If $r _X$ and $r _Y$ are the rate of diffusion of gases $X$ and $Y$,
$\dfrac{r _X}{r _Y} =\dfrac{d}{24-d}=\sqrt{\dfrac{40}{10}}=2 $
$d =48-2 d $
[$\therefore$ Rate of diffusion µ distance travelled]
$\Rightarrow \quad 3 d =48 \Rightarrow \quad d=16 cm$
Hence, (c) is the correct choice.
BETA
