JEE Mains · Chemistry · STD 11 - 1. Some basic concept of chemistry
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 \(8\)
- B \(7\)
- C \(9\)
- D \(10\)
Answer & Solution
Correct Answer
(B) \(7\)
Step-by-step Solution
Detailed explanation
\(\mathrm{M}=\frac{\mathrm{n}_{\mathrm{NaOH}}}{\mathrm{V}_{\mathrm{sol}}(\text { in } \mathrm{L})} \Rightarrow 3=\frac{(84 / 40)}{\mathrm{V}} \Rightarrow \)\(\mathrm{V}=0.7 \mathrm{~L}=7 \times 10^{-1} \mathrm{~L}\)
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