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Q.
How many mL of $ 1\,M $ $ H_{2}SO_{4} $ is required to neutralise $ 10\ mL $ of $ 1\,M $ $ NaOH $ solution?
Chhattisgarh PMTChhattisgarh PMT 2004
Solution:
Molecular weight of $H _{2} SO _{4}=98$
Equivalent weight of $H _{2} SO _{4}=\frac{98}{2}=49$
Molarity $\times$ mol. wt. $=$ normality $\times$ eq. wt.
$1 \times 98=N_{1} \times 49 \,N_{1}=\frac{98}{49}=2$
Similarly, for $N a O H\, 1\, N=1\, M N_{2}=1$
Normality equation, $\underset{\text { Acid }}{N_{1} V_1}=\underset{\text { Base }}{N_{2} V_2}$
$2 \times V_{1}=1 \times 10 V_{1}=5\, m L$
So, $5 \,m L \,1 \,M \,H_{2} S O_{4}$
is required to neutralize $10 \,m L$ of $1 \,M\, NaOH$ solution.