Question

How many kilograms of solvent are there in a sample of ${\text{0}}{\text{.30}}$ molal solution if the sample contains 13 moles of solute?

Answer 1

In the above question, the number of moles of solution and molality of the solution is given. We have to find out the amount of solvent present. This is a direct molality question where we can put the values of the number of moles and molality to get the amount of solvent present.

Formula Used
Molality= $\dfrac{{\text{n}}}{{\text{W}}}$
Where n is the number of moles of solute
W is the weight of the solvent in kg.

Complete step by step solution:
We know that molality can be defined as the number of moles of solute present in one kg of solvent.
In the above question, molality of the solution is given as ${\text{0}}{\text{.30}}$ molal and the number of moles of solute is 13.
Substituting the values in the below equation:
Molality= $\dfrac{{\text{n}}}{{\text{W}}}$, we get:
${\text{0}}{\text{.30 = }}\dfrac{{{\text{13}}}}{{\text{W}}}$
Rearranging the equation, we get that:
${\text{W = }}\dfrac{{{\text{13}}}}{{{\text{0}}{\text{.30}}}}{\text{ = 43}}{\text{.3}}$ kg
Therefore, the weight of the solvent is ${\text{43}}{\text{.3}}$ kg.

Note:
Sometimes, there is a confusion between molarity and molality. In such a case, remember, the difference between molarity and molality lies in the denominator part of the ratio. In molarity, we divide the number of moles by volume of the solution. And in molality, we divide the number of moles by mass of solvent in kg.
Solution=Solute + Solvent
So, in molality solute mass must be subtracted to get the correct result.
In molarity, the unit of the volume of the solution should be taken into consideration and in molality, the weight of solvent be taken into consideration. It should be converted into liters and kg in order to get the correct result.