Question

A solution of two components containing ${\mathrm{n}}_1$moles of the $1^{st}$ component and ${\mathrm{n}}_2$ moles of the second component is prepared. ${\mathrm{M}}_1$and ${\mathrm{M}}_2$ are the molecular weights of component $1$ and $2$ respectively. If $\mathrm{d}$ 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 $2^{nd}$ component, then ${\mathrm{C}}_2$ can be expressed as:

• A. ${\mathrm{C}}_2=\frac{1000{\mathrm{x}}_2}{{\mathrm{M}}_1+{\mathrm{x}}_2\left({\mathrm{M}}_2-{\mathrm{M}}_1\right)}$
• B. ${\mathrm{C}}_2=\frac{\mathrm{d}{\mathrm{x}}_2}{{\mathrm{M}}_2+{\mathrm{x}}_2\left({\mathrm{M}}_2-{\mathrm{M}}_1\right)}$
• C. ${\mathrm{C}}_2=\frac{\mathrm{d}{\mathrm{x}}_1}{{\mathrm{M}}_2+{\mathrm{x}}_2\left({\mathrm{M}}_2-{\mathrm{M}}_1\right)}$
• D. ${\mathrm{C}}_2=\frac{1000\mathrm{d}{\mathrm{x}}_2}{{\mathrm{M}}_1+{\mathrm{x}}_2\left({\mathrm{M}}_2-{\mathrm{M}}_1\right)}$

Answer 1

The correct option is D

${\mathrm{C}}_2=\frac{1000\mathrm{d}{\mathrm{x}}_2}{{\mathrm{M}}_1+{\mathrm{x}}_2\left({\mathrm{M}}_2-{\mathrm{M}}_1\right)}$

1. Given quantities for the 1^st component - number of moles - ${\mathrm{n}}_1$, molecular weight - ${\mathrm{M}}_1$

Given quantities for the 2^nd component (solute) - number of moles - ${\mathrm{n}}_2$, molecular weight - ${\mathrm{M}}_2$, mole fraction - ${\mathrm{x}}_2$

2. Calculation of molarity of 2^nd component $\left({\mathrm{C}}_2\right)$ -

$\mathrm{Molarity}=\frac{\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{mL}}\times 1000—\left(1\right)$

Step 1 - Calculation of moles of 2^nd component-

Let the total number of moles present in the solution be $1$.

${\mathrm{n}}_1+{\mathrm{n}}_2=1$

We know, ${\mathrm{x}}_2=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}$

So, ${\mathrm{x}}_2={\mathrm{n}}_2$ (number of moles of solute / 2^nd component)

and $1-{\mathrm{x}}_2={\mathrm{n}}_1$

Step 2 - Calculation of mass of solution and solute -

Number of moles of solute / 2^nd component $={\mathrm{x}}_2$

So, mass $={\mathrm{x}}_2{\mathrm{M}}_2$ $\mathrm{g}$

Mass of 1^st component $=\left(1-{\mathrm{x}}_2\right)M_1$ $\mathrm{g}$

Mass of the solution $=\left({\mathrm{x}}_2{\mathrm{M}}_2+\left(1-{\mathrm{x}}_2\right){\mathrm{M}}_1\right)$ $\mathrm{g}$

Step 3 - Calculation of volume of the solution -

$\mathrm{volume}=\frac{\mathrm{mass}}{\mathrm{density}}$

Volume of solution $=\frac{\left({\mathrm{x}}_2{\mathrm{M}}_2+\left(1-{\mathrm{x}}_2\right){\mathrm{M}}_1\right)}{\mathrm{d}}$ $\mathrm{mL}$

Step 4- Calculation of molarity of 2^nd component $\left({\mathrm{C}}_2\right)$ - Putting the values of the number of moles of solute and volume of solution in equation 1, we get-${\mathrm{C}}_2=\frac{1000\mathrm{d}{\mathrm{x}}_2}{{\mathrm{M}}_1+{\mathrm{x}}_2\left({\mathrm{M}}_2-{\mathrm{M}}_1\right)}$

Hence, option D is the correct answer.