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Standard XII

Chemistry

Molarity

Number of mil...

Question

Number of mililiters of a $1.6 %$ $B a C l_{2} (w / v)$ solution, which is required to precipitate the sulphur as $B a S O_{4}$ in a $0.60$ g sample that contains $12 %$ $S$ , is:

58.50

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14.63

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29.25

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21.00

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Solution

The correct option is A $29.25$ mL
$S ⟶ B a S O_{4} ⟶ B a C l_{2}$
Therefore, equal moles of sulphur and barium chloride are involved in the reactions.
Weight of sulphur present $= 0.12 \times 0.6 = 0.072$
$B a C l_{2}$ required by $0.072$ g sulphur to be precipitated as $B a S O_{4} = \frac{208}{32} \times 0.072$ $= 0.468$ g.
Thus, volume of $B a C l_{2}$ solution $= \frac{0.468 \times 100}{1.6}$ $= 29.25$ mL.

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Similar questions

Q. Number of millilitres of a

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Number of mililiters of a 1.6% BaCl2(w/v) solution, which is required to precipitate the sulphur as BaSO4 in a 0.60 g sample that contains 12% S, is:

Number of mililiters of a $1.6 %$ $B a C l_{2} (w / v)$ solution, which is required to precipitate the sulphur as $B a S O_{4}$ in a $0.60$ g sample that contains $12 %$ $S$ , is: