Question

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

• A. $C _{2}=\frac{1000 x _{2}}{ M _{1}+ x _{2}\left( M _{2}- M _{1}\right)}$
• B. $C _{2}=\frac{ d x _{2}}{ M _{2}+ x _{2}\left( M _{2}- M _{1}\right)}$
• C. $C _{2}=\frac{ d x _{1}}{ M _{2}+ x _{2}\left( M _{2}- M _{1}\right)}$
• D. $C _{2}=\frac{1000 dx _{2}}{ M _{1}+ x _{2}\left( M _{2}- M _{1}\right)}$

Answer 1

Answer: (D)

$x _{2} \rightarrow$ mole fraction of component $(2) $

$C _{2} \rightarrow$ molarity of component $(2)$

Let total moles of two components $=1 $

Moles of component $-(2)= x _{2}$

Moles of component $-(1)=\left(1-x_{2}\right)$

Mass of component $-(2)= x _{2} M _{2}$

Mass of component $-(1)=\left(1- x _{2}\right) M _{1}$

Total mass $=\left( x _{2} M _{2}+\left(1- x _{2}\right) M _{1}\right) gm$

Total volume $=\left(\frac{ x _{2} M _{2}+\left(1- x _{2}\right) M _{1}}{ d }\right) ml$

Molarity $\left( C _{2}\right)$

$=\frac{\text { Moles of component }(2)}{\text { Volume of solution }(\text { in } ml )} \times 1000$

$=\frac{\frac{ x _{2}}{\frac{ x _{2} M _{2}+\left(1- x _{2}\right) M _{1}}{ d }} \times 1000}{\frac{ d }{ d }}$

$C_{2}=\frac{1000 \cdot d \cdot x _{2}}{ M _{1}+ x _{2}\left( M _{2}- M _{1}\right)}$