States of Matter 1 Question 72
72. (i) One mole of nitrogen gas at $0.8 \mathrm{~atm}$ takes $38 \mathrm{~s}$ to diffuse through a pin-hole, whereas one mole of an unknown compound of xenon with fluorine at $1.6 \mathrm{~atm}$ takes $57 \mathrm{~s}$ to diffuse through the same hole. Calculate the molecular formula of the compound.
(ii) The pressure exerted by $12 \mathrm{~g}$ of an ideal gas at temperature $t^{\circ} \mathrm{C}$ in a vessel of volume $V$ litre is one atm. When the temperature is increased by $10^{\circ} \mathrm{C}$ at the same volume, the pressure increases by $10 %$. Calculate the temperature $t$ and volume $V$.
$($ Molecular weight of the gas $=120)$
(1999, 5M)
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Answer:
Correct Answer: 72. $\left(123 \mathrm{~g} \mathrm{~mol}^{-1}\right)$
Solution:
- (i) For the same amount of gas being effused
$$ \begin{array}{rlrl} & & \frac{r_{1}}{r_{2}} & =\frac{t_{2}}{t_{1}}=\frac{p_{1}}{p_{2}} \sqrt{\frac{M_{2}}{M_{1}}} \ \Rightarrow \quad & \frac{57}{38} & =\frac{0.8}{1.6} \sqrt{\frac{M_{2}}{28}} \ \Rightarrow \quad & M_{2} & =252 \mathrm{~g} \mathrm{~mol}^{-1} \end{array} $$
Also, one molecule of unknown xenon-fluoride contain only one $\mathrm{Xe}$ atom $[M(\mathrm{Xe})=131]$, formula of the unknown gas can be considered to be $\mathrm{XeF}_{n}$.
$\Rightarrow 131+19 n=252 ; n=6.3$, hence the unknown gas is $\mathrm{XeF}_{6}$.
(ii) For a fixed amount and volume, $p \propto T$
$$ \begin{array}{rlrl} \Rightarrow & \frac{1}{1.1} & =\frac{T}{T+10} \quad \text { where, } T=\text { Kelvin temperature } \ \Rightarrow & T & =100 \mathrm{~K}=t+273 \ \Rightarrow & t & =-173^{\circ} \mathrm{C} \ & & \text { Volume } & =\frac{n R T}{p}=\left(\frac{12}{120}\right) \times \frac{0.082 \times 100}{1}=0.82 \mathrm{~L} \end{array} $$
