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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_____________ \(\times 10^{-1} \ \mathrm{dm}^3\).

  1. A \(8\)
  2. B \(7\)
  3. C \(9\)
  4. D \(10\)
Verified Solution

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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