At $25~^\circ\mathrm{C}$, a sample of $\ce{NH}$ effuses at the rate of $0.050$ moles per minute. Under the same conditions, what is the molar mass of a gas that effuses at approximately one half of that rate?

My steps and thoughts

I converted $25~^{\circ}\mathrm{C}$ to $298~\mathrm{K}$ and then I was thinking of using $pV=nRT$ and I got stuck.

I am thinking of using molar mass, but I don't know how to incorporate that.

Please, only give hints not the answer.


1 Answer 1



This question seems to be a simple application of Graham's Law (assuming you know Graham's Law), i.e the rate of effusion of a gas is inversely proportional to the square root of the mass of their particles. This empirical law can be formulated as follows:

$$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{\text{M}_2}{\text{M}_1}}$$


I don't know what the OP meant by $\ce{NH}$, I assume the gas in question is $\ce{NH3}$ with rate of effusion ($R_1$) = 0.050 $\mathrm{mol \min^{-1}}$.
$R_2 = \frac{R_1}{2}$ and $M_1 = 17.031$, plugging all of this in
$M_2 = M_1\cdot \left (\frac{R_1}{R_2} \right )^2 = 4\cdot M_1 = 68.124$


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