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I am stuck on this problem. Suppose we are mixing substances $a$ and $b$, we have

$$\Delta S_{mixing} = -Rn_a \ln (\frac{n_a}{n_a + n_b})-Rn_b \ln (\frac{n_b}{n_a + n_b})$$

and we are told to express $\Delta S_{mixing}$ in terms of the mass ratio $x_a=\frac{m_a}{m_a+m_b}$.

Let $M$ be molar mass, $n$ be moles, $m$ be mass, $R$ be constant, and $x$ the mass ratio.

I have managed to get $$n_a \ln (\frac{n_a}{n_a + n_b})=\frac{-m_a}{M_a}\ln (\frac{M_a m_b}{M_b m_a}+1)$$ and that $$-\ln(\frac{n_a}{n_a +n_b })=- \ln (\frac{n_b}{n_a}+1).$$ The similarity in these two expressions leads me to think I am close to expressing $\Delta S_{mixing}$ in terms of the mass ratio $x_a=\frac{m_a}{m_a+m_b}$ but I have been stuck for a while and am looking for some help.

Note, I do realize that $x_a +x_b =1$ and I suspect this will be useful after I figure out how to express $-Rn_a \ln (\frac{n_a}{n_a + n_b})$ in terms of $x_a$

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    $\begingroup$ Are you saying that you think you can do it without the molecular weights being present.? $\endgroup$ Apr 5, 2019 at 12:07
  • $\begingroup$ No. I just need to get $x_a$ into the $\Delta S_{mixing}$ equation. $\endgroup$ Apr 5, 2019 at 12:11
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    $\begingroup$ Then it's just an algebra problem. $\endgroup$ Apr 5, 2019 at 12:14
  • $\begingroup$ I got stuck so I thought there was maybe something more to it. $\endgroup$ Apr 5, 2019 at 12:20

1 Answer 1

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If you have at total mass m, then the mass of a is $mx_a$ and the number of moles of a is $n_a=\frac{mx_a}{M_a}$. So, $$\frac{n_a}{(n_a+n_b)}=\frac{\frac{x_a}{M_a}}{\left(\frac{x_a}{M_a}+\frac{x_b}{M_b}\right)}=\frac{M_bx_a}{(M_bx_a+M_ax_b)}$$

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