Question

A 500 $c{{m}^{3}}$ solution of anhydrous $N{{a}_{2}}C{{O}_{3}}$ (sodium carbonate) in water contains 26.5 g of it. Calculate molality of the solution if the density of solution is 1 $g/c{{m}^{3}}$ and molecular weight of anhydrous $N{{a}_{2}}C{{O}_{3}}$ = 106 u.

Answer 1

Hint Molality of a solution is nothing but the number of moles of solute is present in 1 kg of solvent. Molality is used to measure the concentration of the solute in the given solution in terms of amount. It is going to be denoted with ‘m’.
$\text{molality(m)=}\dfrac{\text{number of moles of the solute}}{\text{mass of the solvent (in kg)}}$

Complete step by step answer:
- In the question it is given that 26.5 g of sodium carbonate is present in 500 ml of water and asked us to calculate the molality of the solution using the density of the solution.
- Initially we have to calculate the number of moles from the given data.
- The volume of the water is 500 ml.
- The density of the solution is 1g/ml.
- From the above data we can calculate the mass of the solute by using the below formula.
$d=\dfrac{M}{V}$
d = density of the solution = 1 g/ml, M = mass of the solute, V = volume of the solution = 500 ml.
- Substitute the known values in the above formula to get the mass of the solute and it is as follows.

& d=\dfrac{M}{V} \\\ & M=d\times V \\\ & M=1\times 500 \\\ & M=500g \\\ \end{aligned}$$ \- Now mass of the solvent = 500 – 26.5 = 473.5 g. \- Number of moles of the sodium carbonate can be calculated as follows. $$\begin{aligned} & \text{number of moles of sodium carbonate = }\dfrac{\text{Mass of sodium carbonate}}{\text{molecular weight of the sodium carbonate}} \\\ & \implies\dfrac{26.5}{106} \\\ &\implies 0.25moles \\\ \end{aligned}$$ \- Now substitute all the known values in the below formula to get the molality of the solution and it is as follows. $$\begin{aligned} & \text{molality(m)=}\dfrac{\text{number of moles of the solute}}{\text{mass of the solvent (in kg)}} \\\ &\implies \dfrac{0.25}{473.5}\times 1000 \\\ &\implies 0.528m \\\ \end{aligned}$$ **\- Therefore the molality of the solution is 0.528 m.** **Note:** Without knowing the mass of the solute we cannot calculate the molality of the solution. To calculate the mass of the solute we should know the density of the solution. Without knowing the density of the solution we can calculate the mass of the solute.