JEE Mains · Chemistry · STD 11 - 1. Some basic concept of chemistry
\(25\, mL\) of the given \(HCl\) solution requires \(30\, mL\) of \(0.1\, M\) sodium carbonate solution. What is the volume of this \(HCl\) solution required to titrate \(30\, mL\) of \(0.2\, M\) aqueous \(NaOH\) solution? .............. \(\mathrm{mL}\)
- A \(25\)
- B \(75\)
- C \(50\)
- D \(12.5\)
Answer & Solution
Correct Answer
(A) \(25\)
Step-by-step Solution
Detailed explanation
Apply law of equivalence : \(25 \times N = 30 \times 0.1 \times 2\) \({N_{HCl}} = \frac{{30 \times 0.2}}{{25}} = \frac{6}{5} \times 0.2 = \frac{{1.2}}{5}\) For the \(2^{nd}\) titration \(\frac{{1.2}}{5} \times {V_{HCl}} = 30 \times 0.2\)…
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