MCQ
The molality of $1\,M$ solution of $NaCl$ (specific gravity $1.0585\,g/ml$ ) is
- A$1.0585$
- ✓$1$
- C$0.10$
- D$0.0585$
✓
Answer
Correct option: B.
$1$
b
$\mathrm{m}=\frac{\text { mole of } \mathrm{NaCl}}{\text { Weight of solvent in kg. }}=\frac{1}{1}=1$
$\mathrm{m}=\frac{\text { mole of } \mathrm{NaCl}}{\text { Weight of solvent in kg. }}=\frac{1}{1}=1$
weight of solvent $=$ weight of solution $-$ weight of $\mathrm{NaCl}$
${=1.0585 \times 1000-58.5}$
${=1058.5-58.5=1000 \,g=1 \,\mathrm{kg}}$
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