A sample of N_2 gas at 0.8 , atm pressure tak | vedclass

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(B) According to Graham's Law of Effusion,the rate of effusion $r$ is directly proportional to pressure $P$ and inversely proportional to the square r

Vedclass · Accepted Answer

$252$

Vedclass · Answer

(B) According to Graham's Law of Effusion,the rate of effusion $r$ is directly proportional to pressure $P$ and inversely proportional to the square root of molar mass $M$: $r \propto \frac{P}{\sqrt{M}}$.Since $r = \frac{n}{t}$,we have $\frac{n_1}{t_1} \propto \frac{P_1}{\sqrt{M_1}}$ and $\frac{n_2}{t_2} \propto \frac{P_2}{\sqrt{M_2}}$.Given for $N_2$: $P_1 = 0.8 \, atm$,$t_1 = 38 \, s$,$M_1 = 28 \, g/mol$,$n_1 = 1 \, mol$ (assuming equal moles for comparison).Given for gas $X$: $P_2 = 1.6 \, atm$,$t_2 = 57 \, s$,$M_2 = ?$,$n_2 = 1 \, mol$.Using the ratio: $\frac{r_1}{r_2} = \frac{P_1}{P_2} 	imes \sqrt{\frac{M_2}{M_1}} 	imes \frac{t_2}{t_1}$.$\frac{1/38}{1/57} = \frac{0.8}{1.6} 	imes \sqrt{\frac{M_2}{28}} 	imes \frac{57}{38}$.$\frac{57}{38} = 0.5 	imes \sqrt{\frac{M_2}{28}} 	imes \frac{57}{38}$.$1 = 0.5 	imes \sqrt{\frac{M_2}{28}}$.$2 = \sqrt{\frac{M_2}{28}}$.$4 = \frac{M_2}{28}$.$M_2 = 4 	imes 28 = 112 \, g/mol$. (Note: Re-evaluating the calculation: $1.5 = 0.5 	imes \sqrt{M_2/28} 	imes 1.5 \implies 1 = 0.5 	imes \sqrt{M_2/28} \implies 2 = \sqrt{M_2/28} \implies 4 = M_2/28 \implies M_2 = 112$. Since $112$ is not in options,checking the ratio again: $r_1/r_2 = (P_1/P_2) \sqrt{M_2/M_1} = (t_2/t_1)$. $(0.8/1.6) \sqrt{M_2/28} = 57/38 = 1.5$. $0.5 \sqrt{M_2/28} = 1.5 \implies \sqrt{M_2/28} = 3 \implies M_2/28 = 9 \implies M_2 = 252 \, g/mol$.)

$A$ sample of $N_2$ gas at $0.8 \, atm$ pressure takes $38 \, s$ to effuse through a pinhole. $A$ sample of $1 \, mol$ of an unknown gas '$X$' at $1.6 \, atm$ pressure takes $57 \, s$ to effuse through the same pinhole. The molar mass of gas '$X$' in $g/mol$ is .............

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