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1 mL of unknown solution of $H_{2} {SO}_{4}$ is diluted 100 mL and then its 25 mL is titrated with $10 m L$ of $0.2 M NaOH$ . The excess acid required $10 mL of 0.1 s$ $Ba {(OH)}_{2}$ solution. Find molarity of original $H_{2} {SO}_{4}$ solution.

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answer is 8.

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Detailed Solution

Here,

$25 mI$ of $H_{2} {SO}_{4}$ contains milli equivalent

=10 X 0.2 + 10 X 0.1 X 2 = 4

$\begin{array}{l} ∴ 100 ml contains = 16 \\ 1 \times M \times 2 = 16 \\ \Rightarrow M = 8 \end{array}$

Thus, the required molarity is 8.

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