A solution of two components containing $n_{1}$ moles of the 1st component and $n_{2}$ moles of the $2^{n d}$ 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 $g m L^{- 1}$ , $C_{2}$ is the molarity and $X_{2}$ is the mole fraction of the 2nd component, then $C_{2}$ can be expressed as

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$C_{2} = \frac{{dx}_{1}}{M_{2} + x_{2} (M_{2} - M_{1})}$

$C_{2} = \frac{{dx}_{2}}{M_{2} + x_{2} (M_{2} - M_{1})}$

$C_{2} = \frac{1000 {dx}_{2}}{M_{1} + x_{2} (M_{2} - M_{1})}$

$C_{2} = \frac{1000 x_{2}}{M_{1} + x_{2} (M_{2} - M_{1})}$

answer is C.

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Detailed Solution

$C a l c u l a t i o n o f t h e m o l e s o f t h e s e c o n d c o m p o n e n t : A s s u m e t h a t t h e t o t a l n u m b e r o f m o l e s i n t h e s o l u t i o n i s 1 . n_{1} + n_{2} = 1 A s w e k n o w, x_{2} = \frac{n_{2}}{n_{1} + n_{2}} S o, x_{2} = n_{2} (n u m b e r o f m o l e s o f s o l u t e / 2 n d c o m p o n e n t) a n d 1 - x_{2} = n_{1} Calculation of mass of solution and solute - Number of moles of the second component / solvent So, mass = x_{2} M_{2} g Mass of 1 st component = (1 - x_{2}) M_{1} g Mass of solution = (x_{2} M_{2} + (1 - x_{2}) M_{1}) g Calculation of the volume of the solution - volume = \frac{mass}{density} Volume of solution = \frac{(x_{2} M_{2} + (1 - x_{2}) M_{1})}{d} mL Calculation of the molarity of component (C_{2}) we put the values of the molar amount of the solute and the volume of the solution into equation 1, we get - C_{2} = \frac{1000 {dx}_{2}}{M_{1} + x_{2} (M_{2} - M_{1})}$