JEE Mains · Chemistry · STD 11 - 1. Some basic concept of chemistry
\(2\,L\) of \(0.2\,M H _{2} SO _{4}\) is reacted with \(2\,L\) of \(0.1\,M\) \(NaOH\) solution, the molarity of the resulting product \(Na _{2} SO _{4}\) in the solution is millimolar. (Nearest integer).
- A \(24\)
- B \(23\)
- C \(22\)
- D \(25\)
Answer & Solution
Correct Answer
(D) \(25\)
Step-by-step Solution
Detailed explanation
\(H _{2} SO _{4}+2 NaOH \rightarrow Na _{2} SO _{4}+2 H _{2} O\) \(0.4\,mol \quad 0.2 mol \quad-\) \(0.3\,mol \quad-\quad\quad\quad\quad\quad 0.1 mol\) Molarity of \(Na _{2} SO _{4}\) is \(\frac{0.1}{4}=0.025\,M\) \(=25\,mM\).
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