# The molarity M of a solution is related to the mole fraction x2 of solute by the expression, where p is the density of solution and M1 and M2 are the molar masses of solvent and solute, respectively.

1. A

$\begin{array}{r}M=\frac{{x}_{2}\rho }{{M}_{1}+{x}_{2}\left({M}_{1}-{M}_{2}\right)}\end{array}$

2. B

$\begin{array}{r}M=\frac{{x}_{2}\rho }{{M}_{1}-{x}_{2}\left({M}_{1}-{M}_{2}\right)}\end{array}$

3. C

$\begin{array}{r}M=\frac{{x}_{2}\rho }{{M}_{1}+{x}_{2}\left({M}_{1}+{M}_{2}\right)}\end{array}$

4. D

$\begin{array}{r}M=\frac{{x}_{2}\rho }{{M}_{1}-{x}_{2}\left({M}_{1}+{M}_{2}\right)}\end{array}$

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### Solution:

Suppose that the l and 2 denote solvent and solute respectively,

$\begin{array}{l}{\mathrm{x}}_{2}=\frac{{\mathrm{n}}_{2}}{{\mathrm{n}}_{1}+{\mathrm{n}}_{2}}=\frac{{\mathrm{n}}_{2}}{\frac{{\mathrm{m}}_{1}}{{\mathrm{M}}_{1}}+{\mathrm{n}}_{2}}\\ =\frac{{\mathrm{n}}_{2}{\mathrm{M}}_{1}}{{\mathrm{m}}_{1}+{\mathrm{n}}_{2}{\mathrm{M}}_{1}}=\frac{{\mathrm{n}}_{2}{\mathrm{M}}_{1}}{\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)+{\mathrm{n}}_{2}{\mathrm{M}}_{1}-{\mathrm{m}}_{2}}\\ =\frac{{\mathrm{n}}_{2}{\mathrm{M}}_{1}}{{\mathrm{V}}_{\mathrm{e}}+{\mathrm{n}}_{2}{\mathrm{M}}_{1}-{\mathrm{m}}_{2}}=\frac{{\mathrm{n}}_{2}{\mathrm{M}}_{1}}{{\mathrm{V}}_{\mathrm{p}}+{\mathrm{n}}_{2}\left({\mathrm{M}}_{1}-{\mathrm{M}}_{2}\right)}\\ =\frac{\left({\mathrm{n}}_{2}/\mathrm{v}\right){\mathrm{M}}_{1}}{\mathrm{p}+\left({\mathrm{n}}_{2}/\mathrm{v}\right)\left({\mathrm{M}}_{1}-{\mathrm{M}}_{2}\right)}=\frac{{\mathrm{MM}}_{1}}{\mathrm{p}+\mathrm{M}\left({\mathrm{M}}_{1}-{\mathrm{M}}_{2}\right)}\end{array}$

Thus

$\begin{array}{r}M=\frac{{x}_{2}\rho }{{M}_{1}-{x}_{2}\left({M}_{1}-{M}_{2}\right)}\end{array}$.

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