# A solution of two components containing n1 moles of the 1st

Question:

A solution of two components containing $\mathrm{n}_{1}$ moles of the $1^{\text {st }}$ component and $\mathrm{n}_{2}$ moles of the $2^{\text {nd }}$ component is prepared. $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$ are the molecular weights of component 1 and 2 respectively, If $d$ is the density of the solution in $\mathrm{g} \mathrm{mL}^{-1}, \mathrm{C}_{2}$ is the molarity and $\mathrm{x}_{2}$ is the mole fraction of the $2^{\text {nd }}$ component, then $C_{2}$ can be expressed as :

1. $\mathrm{C}_{2}=\frac{1000 \mathrm{x}_{2}}{\mathrm{M}_{1}+\mathrm{x}_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}$

2. $\mathrm{C}_{2}=\frac{\mathrm{dx}_{2}}{\mathrm{M}_{2}+\mathrm{x}_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}$

3. $\mathrm{C}_{2}=\frac{\mathrm{dx}_{1}}{\mathrm{M}_{2}+\mathrm{x}_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}$

4. $\mathrm{C}_{2}=\frac{1000 \mathrm{dx}_{2}}{\mathrm{M}_{1}+\mathrm{x}_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}$

Correct Option: , 4

Solution:

(4) $\mathrm{C}_{2}=\frac{1000 \mathrm{dx}_{2}}{\mathrm{M}_{1}+\mathrm{x}_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}$