Question

An antifreeze solution is prepared from $222.6g$ of ethylene glycol $\left(C_2{H}_6{O}_2\right)$ and $200g$ of water. Calculate the molality of the solution. If the density of the solution is $1.072gm{L}^{-1}$, then what shall be the molarity of the solution?

Answer 1

Step 1: Calculating the molality of the solution

Since ethylene glycol is present in excess, it acts as a solvent and water is a solute.

The molar mass of water is $18g/mol$

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Number of moles of solute $=\frac{200}{18}=11.11$ moles

Molality is defined as the number of moles of solute present in the $1$ kg of solvent.

$\text{Molality}=\frac{\text{Number of moles of solute}}{\text{mass of solvent(g)}}\times 1000$

$\text{Molality}=\frac{11.11}{222.6}\times 1000=49.9$ m

Step 2: Calculating the molarity of the solution

The molarity of a solution is defined as the number of moles of solute dissolved in 1 Litre of solution.

$\text{Molarity}=\frac{\text{moles of solute}}{\text{Volume of solution(mL)}}\times 1000$

$\text{Total mass of solution}=222.6+200=422.6g$

$\text{Density of solution}=1.072g/mL$

$\text{Volume of solution}=\frac{\text{Mass}}{\text{Density}}=\frac{422.6}{1.072}=394mL$

$\text{Molarity}=\frac{11.11}{0.394}\times 1000=28.2M$