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Standard XII

Chemistry

Percentage Composition

The weight of...

Question

The weight of mixture containing HCl and $H_{2} S O_{4}$ is $0.1$ g on treatment with an excess of an $A g N O_{3}$ solution, reacted with this acid mixture gives $0.1435$ g of AgCl, weight percentage of $H_{2} S O_{4}$ in mixture is:

36.5

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63.5

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50

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80

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Solution

The correct option is A $63.5$
Let mass of $H C l$ be $x$
$∴$ Mass of $H_{2} S O_{4}$ is $0.1 - x$
Moles of $H C l = \frac{x}{36.5}$
$H C l + A g N O_{3} ⟶ A g C l ↓ + H N O_{3}$
Moles of $H C l =$ moles of $A g C l$
Moles of $A g C l = \frac{0.1435}{143.3} = 10^{- 3}$ moles
$∴ \frac{x}{36.5} = 10^{- 3}$
$x = 36.5 \times 10^{- 3}$
Weight of $H C l = 0.0365$ $g$
Weight of $H_{2} S O_{4} = 0.0635$
$%$ weight of $H_{2} S O_{4} = \frac{0.0635}{0.1} \times 100 = 63.5 %$

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The weight of mixture containing HCl and H2SO4 is 0.1g on treatment with an excess of an AgNO3 solution, reacted with this acid mixture gives 0.1435g of AgCl, weight percentage of H2SO4 in mixture is:

The correct option is A 63.5Let mass of HCl be x∴ Mass of H2SO4 is 0.1−xMoles of HCl=x36.5HCl+AgNO3⟶AgCl↓+HNO3Moles of HCl= moles of AgClMoles of AgCl=0.1435143.3=10−3 moles∴x36.5=10−3x=36.5×10−3Weight of HCl=0.0365 gWeight of H2SO4=0.0635% weight of H2SO4=0.06350.1×100=63.5%

The weight of mixture containing HCl and $H_{2} S O_{4}$ is $0.1$ g on treatment with an excess of an $A g N O_{3}$ solution, reacted with this acid mixture gives $0.1435$ g of AgCl, weight percentage of $H_{2} S O_{4}$ in mixture is: