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 $gm{L}^{-1},C_2$ is the molarity and $x_2$ is the mole fraction of the $2^{\text{nd }}$ component, then $C_2$ can be expressed as :

• A. $C_2=\frac{1000x_2}{M_1+x_2\left(M_2-M_1\right)}$
• 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. $C_2=\frac{1000d{x}_2}{M_1+x_2\left(M_2-M_1\right)}$

Answer 1

Answer: (D)

$x_2\to$ mole fraction of component $(2)$

$C_2\to$ 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 $\left(C_2\right)$

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

$=\frac{\frac{x_2}{\frac{x_2{M}_2+\left(1-x_2\right)M_1}{d}}\times 1000}{\frac{d}{d}}$

$C_2=\frac{1000\cdot d\cdot x_2}{M_1+x_2\left(M_2-M_1\right)}$