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Standard XII

Chemistry

Stoichiometric Calculations

10 g sample o...

Question

$10$ g sample of bleaching powder was dissolved into water to make the solution one litre. To this solution, $35$ mL of $1.0$ M Mohr salt solution was added containing enough $H_{2} S O_{4}$ . After the reaction was complete, the excess Mohr salt required $30$ mL of $0.1$ $M K M n O_{4}$ for oxidation. The $%$ of available $C l_{2}$ approximately is (Molecular weight $= 71$ g/mol) :

7.1 %

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8.1 %

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9.1 %

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6.1 %

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Solution

The correct option is C $7.1 %$
Meq of Mohr's salt
$= 35 \times 1 \times 1 = 35$
M eq of $K M n O_{4} =$ Meq of excess Mohr salt
$= 30 \times 0.1 \times 5 = 15$
M.eq of Mohr salt reacted with bleaching powder $= 35 - 15 = 20$
M eq of $C l_{2} = 20$
Weight of $C l_{2} = 20 \times 10^{- 3} \times \frac{71}{2} = 0.71 g$
Percent of $C l_{2} = \frac{0.71}{10} \times 100 = 7.1$ %

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The correct option is C 7.1% Meq of Mohr's salt =35×1×1=35M eq of KMnO4= Meq of excess Mohr salt =30×0.1×5=15M.eq of Mohr salt reacted with bleaching powder =35−15=20M eq of Cl2=20Weight of Cl2=20×10−3×712=0.71gPercent of Cl2=0.7110×100=7.1%