In an experiment conducted to study the diffusion of gases using … - Vidyadip

Question

In an experiment conducted to study the diffusion of gases using same experimental conditions, following data were recorded.
Gas $A: 50 \ cm^3$ of gas $A$ takes $7$ minutes to diffuse from one container to the adjacent container.
Gas $B: 50 \ cm^3 $of gas $B$ takes $10$ minutes to diffuse from one container to the adjacent container.
$i.$ What is the rate of diffusion of gas $A$?
$ii.$ What is the rate of diffusion of gas $B$?
$iii.$ Which gas has higher molecular mass?

✓

Answer

$i.$ Volume of gas $A$ diffused $=50 \ cm ^3$
Time required for diffusion $=7$ minutes $=7 \times 60$ seconds
Rate of diffusion of a gas $=\frac{\text { Volume of the gas diffused }}{\text { Time required for diffusion }}=\frac{50 cm ^3}{7 \times 60 s }= 0 . 1 2 ~ c m ^{ 3 } ~ s ^{-1}$
$\therefore$ The rate of diffusion of gas $A$ is $0.12 \ cm ^3 s ^{-1}$.
$ii.$ The rate of diffusion of gas $B$ is $0.083 \ cm ^3 s ^{-1}$.
$iii.$ Gas $B$ has higher molecular mass.

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