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 : \(\mathrm{H}: 1.0, \mathrm{C}: 12.0, \mathrm{O}: 16.0]\)
- A \(0.20\)
- B \(2\)
- C \(200\)
- D \(20\)
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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