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JEE Mains · Chemistry · STD 11 - 1. Some basic concept of chemistry

The molarity of the solution prepared by dissolving \(6.3\, \mathrm{~g}\) of oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right)\) in \(250\, \mathrm{~mL}\) of water in \(\operatorname{mol} \mathrm{L}^{-1}\) is \(\mathrm{x} \times 10^{-2} .\) The value of \(\mathrm{x}\) is ..... . (Nearest integer) [Atomic mass : \(\mathrm{H}: 1.0, \mathrm{C}: 12.0, \mathrm{O}: 16.0]\)

  1. A \(0.20\)
  2. B \(2\)
  3. C \(200\)
  4. D \(20\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(20\)

Step-by-step Solution

Detailed explanation

\({\left[\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right]=\frac{\text { weight } / \mathrm{M}_{\mathrm{w}}}{\mathrm{V}(\mathrm{L})}}\) \(\Rightarrow \mathrm{x} \times 10^{-2}=\frac{6.3 / 126}{250 / 1000}\) \(\mathrm{x}=20\)
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