Question

Calculate the molality of a $1L$ solution of $68\%H_{2}S{O}_4\left(\frac{w}{v}\right),$ if the density of the solution is given as $1.80gm{L}^{-1}$.

• A. $6.8m$
• B. $6.2m$
• C. $6.94m$
• D. $10.8m$

Answer 1

The correct option is B $6.2m$

68% $\left(\frac{w}{v}\right)H_{2}S{O}_4=68g{H}_{2}S{O}_4$ in $100mL$ solution.
i.e., $680g$ $H_{2}S{O}_4$ in $1000mL$ solution.
Mass of solution $=$ volume $\times$ density
$=1000mL\times 1.80g/mL$
$=1800g$
Mass of solvent = $1800–680=1120g$
Again, Molar mass of $H_{2}S{O}_4=2\times 1+32+16\times 4=98g$
Number of moles of solute,
$\frac{\text{Mass}}{\text{Molar mass}}=\frac{680}{98}=6.94$ mol
Molality:
$=\frac{\text{No. of moles of solute}}{\text{Mass of solvent in kg}}=\frac{(6.94\times 1000)}{1120}=6.2m$