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.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\begin{aligned} M = \frac{x_{2} ρ}{M_{1} + x_{2} (M_{1} - M_{2})} \end{aligned}$

$\begin{aligned} M = \frac{x_{2} ρ}{M_{1} - x_{2} (M_{1} - M_{2})} \end{aligned}$

$\begin{aligned} M = \frac{x_{2} ρ}{M_{1} + x_{2} (M_{1} + M_{2})} \end{aligned}$

$\begin{aligned} M = \frac{x_{2} ρ}{M_{1} - x_{2} (M_{1} + M_{2})} \end{aligned}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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

$\begin{array}{l} x_{2} = \frac{n_{2}}{n_{1} + n_{2}} = \frac{n_{2}}{\frac{m_{1}}{M_{1}} + n_{2}} \\ = \frac{n_{2} M_{1}}{m_{1} + n_{2} M_{1}} = \frac{n_{2} M_{1}}{(m_{1} + m_{2}) + n_{2} M_{1} - m_{2}} \\ = \frac{n_{2} M_{1}}{V_{e} + n_{2} M_{1} - m_{2}} = \frac{n_{2} M_{1}}{V_{p} + n_{2} (M_{1} - M_{2})} \\ = \frac{(n_{2} / v) M_{1}}{p + (n_{2} / v) (M_{1} - M_{2})} = \frac{{MM}_{1}}{p + M (M_{1} - M_{2})} \end{array}$