MCQ
Volume of $3 \ \mathrm{M} \ \mathrm{NaOH}$ (formula weight $40 \mathrm{~g} \mathrm{~mol}^{-1}$ ) which can be prepared from $84 \mathrm{~g}$ of$\mathrm{NaOH}$ is_____________ $\times 10^{-1} \ \mathrm{dm}^3$.
- A$8$
- ✓$7$
- C$9$
- D$10$
✓
Answer
Correct option: B.
$7$
b
$\mathrm{M}=\frac{\mathrm{n}_{\mathrm{NaOH}}}{\mathrm{V}_{\mathrm{sol}}(\text { in } \mathrm{L})} \Rightarrow 3=\frac{(84 / 40)}{\mathrm{V}} \Rightarrow \mathrm{V}=0.7 \mathrm{~L}=7 \times 10^{-1} \mathrm{~L}$
$\mathrm{M}=\frac{\mathrm{n}_{\mathrm{NaOH}}}{\mathrm{V}_{\mathrm{sol}}(\text { in } \mathrm{L})} \Rightarrow 3=\frac{(84 / 40)}{\mathrm{V}} \Rightarrow \mathrm{V}=0.7 \mathrm{~L}=7 \times 10^{-1} \mathrm{~L}$
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