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(C) According to Graham's law of diffusion for gases at different pressures,the rate of diffusion $r$ is given by $r \propto \frac{P}{\sqrt{M}}$.Thus,

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$2.198$

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(C) According to Graham's law of diffusion for gases at different pressures,the rate of diffusion $r$ is given by $r \propto \frac{P}{\sqrt{M}}$.Thus,$\frac{r_{HCl}}{r_{NH_3}} = \frac{P_{HCl}}{P_{NH_3}} \sqrt{\frac{M_{NH_3}}{M_{HCl}}}$.Given that $NH_4Cl$ forms at $60 \, cm$ from the $HCl$ end,the distance traveled by $HCl$ is $60 \, cm$ and by $NH_3$ is $40 \, cm$. Since the rate is proportional to the distance traveled in the same time,$\frac{r_{HCl}}{r_{NH_3}} = \frac{60}{40} = 1.5$.Substituting the values: $1.5 = \frac{P}{1} \sqrt{\frac{17}{36.5}}$.$P = 1.5 	imes \sqrt{\frac{36.5}{17}} = 1.5 	imes \sqrt{2.147} \approx 1.5 	imes 1.465 = 2.198 \, atm$.

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