Question

Find the molality of 15% (w/w) solution of $H_{2}S{O}_4$.

• A. 1.09 m
• B. 1.64 m
• C. 1.80 m
• D. 2.05 m

Answer 1

The correct option is C 1.80 m

15 % by mass $H_{2}S{O}_4$ means 15 g is present in 100 g of water
Number of moles of $H_{2}S{O}_4=\frac{\text{given mass}}{\text{molar mass}}=\frac{15}{98}=0.153$
Weight of solvent = 100 – 15 = 85 g
Molality $=\frac{\text{moles of solute}}{\text{Weight of solvent}\left(g\right)}\times 1000$
Molality $=\frac{\text{0.153}}{\text{85}}\times 1000$ = 1.8 m