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20ml of 1M ...

Question

$20$ ml of $1 M B a C l_{2}$ solution and $25$ ml of $0.1$ M $N a_{2} S O_{4}$ are mixed to form $B a S O_{4}$ . The amount of $B a S O_{4}$ precipitated are:

A

2.5

g

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B

2.5

moles

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C

2.5

millimoles

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D

0.05

millimoles

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Solution

The correct option is C $2.5$ millimoles
$B a C l_{2} + N a_{2} S O_{4} ⟶ B a S O_{4} ↓ + 2 N a C l$
$1$ mole of $B a C l_{2}$ and $1$ mole of $N a_{2} S O_{4}$ mix to form $1$ mole of $B a S O_{4}$
Moles present of $B a C l_{2} = 20 \times 1 \times 10^{- 3} = 20 m m o l e s$ .
Moles of $N a_{2} S O_{4} = 25 \times 0.1 \times 10^{- 3} = 2.5 m m o l e s .$
$∴$ Moles of $B a S O_{4}$ ppt $= 2.5 m m o l e s .$

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