Question

A compound ${{H}_{2}}X$ with molar weight of 80g is dissolved in a solvent having density of 0.4g$m{{l}^{-1}}$. Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is:
(A)- 6
(B)- 7
(C)- 8
(D)- 9

Answer 1

Hint: Molality of a solution is the number of moles of solute dissolved in one kg of the solvent. It is expressed in moles \[k{{g}^{-1}}\]. It is calculated as
     \[\text{Molality (m) = }\dfrac{\text{no}\text{. of moles of solute}}{\text{weight of solvent (in kg)}}=\dfrac{\text{no}\text{. of moles of solute}}{\text{weight of solvent (in g)}}\times 1000\]
Molarity of a solution is equal to the number of moles dissolved per liter of the solution.
     \[\]

Complete answer:
According to the question, we have
Molar weight of the compound ${{H}_{2}}X$= 80 g
Density of the solvent = 0.4g$m{{l}^{-1}}$
Molarity of the solution = 3.2 M
3.2 molar solution means that there are 3.2 moles of the compound ${{H}_{2}}X$dissolved in one liter of the solution.
Therefore, from 3.2 M solution we have,
Number of moles of the solute ${{H}_{2}}X$= 3.2
Volume of the solution = 1 liter
 Now, we have been given that no change in the volume of the solution occurs when the solute is dissolved in the solvent, i.e. $\text{Volum}{{\text{e}}_{\text{solution}}}\text{=Volum}{{\text{e}}_{\text{solvent}}}$= 1 liter.
Thus, volume of the solvent = 1 liter.
We know that density is given as
     \[\text{Density = }\dfrac{\text{mass}}{\text{volume}}\]
The density of the solvent is 0.4g$m{{l}^{-1}}$.
Substituting volume of the solvent, $\text{Volum}{{\text{e}}_{\text{solvent}}}$ = 1 liter = 1000 ml and its density = 0.4g$m{{l}^{-1}}$ in the above equation, we can find the mass of the solvent
Mass of the solvent = density of the solvent $\times $ volume of the solvent
Mass of the solvent = 0.4g$m{{l}^{-1}}$ $\times $ 1000 ml = 400 g = 0.4 kg
Therefore, the molality of the solution will be
     \[\text{Molality (m) = }\dfrac{\text{no}\text{. of moles of solute (}{{\text{H}}_{2}}\text{X)}}{\text{weight of solvent (in kg)}}\]
Substituting the number of moles = 3.2 mol and mass of the solvent = 0.4 kg in the molality expression, we get
     \[m=\text{ }\dfrac{3.2\,mol}{0.4\,kg}=\dfrac{32\,mol}{4\,kg}=8mol\,k{{g}^{-1}}\]
Therefore, the molality of the solution is 8 moles $k{{g}^{-1}}$.

Hence, the correct option is (C).

Note:
Do not get confused between molality and molarity. Molality does not involve the volume but mass of the solvent. Molarity involves the volume of the solution. Make use of the assumption to reach the solution.