JEE Mains · Chemistry · STD 11 - 1. Some basic concept of chemistry
\(50\, mL\) of \(0.5\, M\) oxalic acid is needed to neutralize \(25\, ml\) of sodium hydroxide solution. The amount ....... gram of \(NaOH\) in \(50\, mL\) of the given sodium hydroxide solution is
- A \(40\)
- B \(10\)
- C \(20\)
- D None of these
Answer & Solution
Correct Answer
(A) \(40\)
Step-by-step Solution
Detailed explanation
Eq. of \((COOH)_2\) \(=\) Eq. of \(NaOH\) \(50 \times 0.5 \times 2 = 25 \times M \times 1\) Mass of \(NaOH\) in \(50\, mL\) \( = \frac{{50 \times 2}}{{1000}} \times 40 = 4\,g\)
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