A $1.00~\mathrm{L}$ gas at $100~^\circ\mathrm{C}$ and $500~\mathrm{Torr}$ contains $70.0\ \%~\ce{He}$ and $30.0\ \%~\ce{Ne}$ by mass. What is the partial pressure of $\ce{He}$?

My answer is $350~\mathrm{Torr}$ whereas the correct answer is $421~\mathrm{Torr}$.

Here is my solution (source; commas , should be points .; I used to write it this way.)

$$\begin{aligned} n_\mathrm{total} &= \frac{pV}{RT} = \frac{(1\ \mathrm l)(0.652\ \mathrm{atm})}{(0.082\ \mathrm{l\ atm\ K^{-1}\ mol^{-1}})(373\ \mathrm K)}=0.0215~\mathrm{mol}\\ 70\ \%~\ce{He} &\implies (70\ \%)(0.0215\ \mathrm{mol}) = 0.0151~\mathrm{mol}~\ce{He}\\ p_\ce{He} &= \frac{nRT}{V}\\ &= \frac{(0.0151\ \mathrm{mol})(0.082\ \mathrm{l\ atm\ K^{-1}\ mol^{-1}})(373\ \mathrm K)}{1\ \mathrm l} = 0.460~\mathrm{atm}\end{aligned}$$

$$0.460~\mathrm{atm}\cdot\frac{101325~\mathrm{Pa}}{1~\mathrm{atm}}\cdot\frac{1~\mathrm{Torr}}{133.322~\mathrm{Pa}}\approx 350~\mathrm{Torr}$$


You're getting a different answer because partial pressure is proportional to the fraction of each gas in moles, rather than by weight. Why is that? Looking at $pV=nRT$, you can see that the ratio between partial pressures is the same as the ratio between number of moles (since $V, R, T$ are the same for both gases).

So, if you respect proportions, your answer will come out the same whether you consider $1.00\ \mathrm{L}$ of the mix or some unspecified number of liters. This lets you avoid a messy calculation with $pV/RT$. You may as well assume that there are 70 g of helium and 30 g of neon. Find how many moles of each gas that represents. The partial pressures of each gas are in the same proportion as the number of moles. With this and the total pressure, you'll get the answer.

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