Question

For a solution of density, d in g/ml containing solute of molecular weight W, the molarity M and molality m are related by:
A. $\dfrac{d}{M} = \dfrac{1}{m} + \dfrac{{M'}}{{1000}}$
B. $\dfrac{d}{m} = \dfrac{1}{M} + \dfrac{W}{{1000}}$
C. $\dfrac{W}{M} = \dfrac{1}{m} + \dfrac{d}{{1000}}$
D. $\dfrac{d}{m} = \dfrac{W}{M} + \dfrac{1}{{1000}}$

Answer 1

Write the expressions for the molarity, molality and density. Rearrange these expressions. From the expression for molarity, write a rearranged expression for volume. From the expression for molality, write a rearranged expression for the weight of solvent. Then work with these rearranged expressions and derive a relationship between the molarity M and molality m.

Complete Step by step answer: Assume that W is the mass of solute, V is the volume of solution, m and M are the molality and molarity respectively. Also assume that M' is the molar mass of solute and W' is the weight of the solvent.
Write the expressions for the molarity, molality and density.
Molarity ${\text{M = }}\dfrac{W}{{M'}} \times \dfrac{{1000}}{V}$ ... ...(1)
Molality ${\text{m = }}\dfrac{W}{{M'}} \times \dfrac{{1000}}{{W'}}$... ... (2)
Density ${\text{d = }}\dfrac{{{\text{mass}}}}{V}{\text{ = }}\dfrac{{\left( {{\text{W + W'}}} \right)}}{V}$
From the expression for molality (equation (1)), write a rearranged expression for weight of solvent.
$V = \dfrac{{W \times 1000}}{{MM'}}$ ... ...(3)
From the expression for molality (equation (2)), write a rearranged expression for weight of solvent.
${\text{W' = }}\dfrac{{{\text{W}} \times {\text{1000}}}}{{{\text{mM'}}}}$
Add W on both sides of the equation.
$W + W' = W + \dfrac{{W \times 1000}}{{mM'}}$
On the right hand side, take $\dfrac{{1000}}{{M'}}W$ as common factor
$W + W' = \dfrac{{1000}}{{M'}}\left[ {\dfrac{{M'}}{{1000}} + \dfrac{1}{m}} \right]W$... ...(4)
Divide equation (4) with equation (3) to obtain the following

$$\Rightarrow \dfrac{{W + W'}}{V} = \dfrac{{\dfrac{{1000}}{{M'}}\left[ {\dfrac{{M'}}{{1000}} + \dfrac{1}{m}} \right]W}}{{\dfrac{{W \times 1000}}{{MM'}}}} \\\ \Rightarrow d = \dfrac{{W + W'}}{V} = \left[ {\dfrac{{MM'}}{{1000}} + \dfrac{M}{m}} \right] \\\$$

Divide above equation with M
$\dfrac{d}{M} = \dfrac{{M'}}{{1000}} + \dfrac{1}{m}$
$\Rightarrow \dfrac{d}{M} = \dfrac{1}{m} + \dfrac{{M'}}{{1000}}$
The molarity M and molality m are related by the expression $\dfrac{d}{M} = \dfrac{1}{m} + \dfrac{{M'}}{{1000}}$

Hence, the correct option is the option (A).

Note: Molarity is the number of moles of solute present in one liter of solution. Molality is the number of moles of solute present in one kilogram of solvent. Density is the ratio of the mass (in grams) to volume (in liters).