This question is based on Graham's law of diffusion - "The diffusion rate of a gas is inversely proportional to the square root of its density". Comparing two gases, the expression is:
\(\frac{r_{1}}{r_{2}}=\sqrt{\frac{\rho_{2}}{\rho_{1}}}=\sqrt{\frac{M_{2}}{M_{1}}}\)
where \(\rho_{1}, \rho_{2}\) and \(M_{1}, M_{2}\) represent the densities and the relative molecular masses of the gases involved. The fastest approach to this question is to admit that the diffusion rate of a gas is directly proportional to the distance travelled. We can therefore replace the ratio of the diffusion rates with the ratio of the distances travelled. Hence, \(\frac{\text { distance travelled by } NH _{3}}{\text { distance travelled by } HCl }\)
\begin{aligned}
&=\sqrt{\frac{M_{M C l}}{M_{ NH _{3}}}}=\sqrt{\frac{36.5}{17}} \\
&=1.465
\end{aligned}
We can use the direct parameters, which is however less time-saving:iffusion rate \(=\frac{\text { Distance travelled }}{\text { time taken }}\)
let the distance travelled by ammonia be \(x\). Hence, the distance travelled by \(HCl\)
will be \(100-x\) (since the tube is \(100 cm\) long)
\begin{aligned}
r_{N H}, & \frac{x}{t}, r_{H C l} \Rightarrow \frac{100-x}{t} \\
\frac{r_{H C l}}{r_{N I_{3}}} &=\sqrt{\frac{M_{N H_{3}}}{M_{H C l}}} \\
\frac{100-x}{t} \div \frac{x}{t} &=\sqrt{\frac{17}{36.5}} \\
\frac{100-x}{t} \times \frac{t}{x} &=0.6825 \\
\frac{100-x}{x} &=0.6825 \\
100-x &=0.6825 x \\6825 x &=100 \\
\text { Distance travelled by } & \\
\text { HCl }=100-x &=(100-59.435) cm \\
=40.565 cm
\end{aligned}
The ratio of distance covered by ammonia to that covered by \(HCl\) is thus \(59.435 cm / 40.565 cm =1.465\) as before.