Change of entropy of mixing in terms mass ratio

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 $$x_a$$

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

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, $$\frac{n_a}{(n_a+n_b)}=\frac{\frac{x_a}{M_a}}{\left(\frac{x_a}{M_a}+\frac{x_b}{M_b}\right)}=\frac{M_bx_a}{(M_bx_a+M_ax_b)}$$