MCQ
Density of a $2.05\,M$ solution of acetic acid in water is $1.02\, g/mL.$ The molality of the solution is ............. $\mathrm{mol}$ $\mathrm{kg}^{-1}$
- ✓$2.28$
- B$0.44$
- C$1.14$
- D$3.28$
✓
Answer
Correct option: A.
$2.28$
a
Molality $(\mathrm{m})=\frac{1000 \times M}{1000 \times d-M \times M_{\text {sdute }}}$
Molality $(\mathrm{m})=\frac{1000 \times M}{1000 \times d-M \times M_{\text {sdute }}}$
Here $M=$ molarity, $M$ solute $=$ molecular mass of solute, $d=$ density of solution
$\therefore m=\frac{1000 \times 2.05}{1000 \times 1.02-2.05 \times 00}=2.28$
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