JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
The normality of \(H _{2} SO _{4}\) in the solution obtained on mixing \(100 mL\) of \(0.1\,M H _{2} SO _{4}\) with \(50 mL\) of \(0.1\) \(M\,NaOH\) is \(\times 10^{-1}\,N\). (Nearest Integer)
- A \(1\)
- B \(2\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
No. of equivalents of \(H _{2} SO _{4}=100 \times 0.1 \times 2=20\) No. of equivalents of \(NaOH =50 \times 0.1=5\) No. of equivalents of \(H _{2} SO _{4}\) left \(=20-5=15\) \(150 \times x =15\) \(x =\frac{1}{10}=0.1 N =1 \times 10^{-1}\,N\)
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