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Question

AgNO3 sample is 73% by mass. To prepare 125 mL of 0.04 M AgNO3 solution, AgNO3 sample required is

A
1.471 g
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B
1.164 g
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C
1.250 g
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D
1.063 g
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Solution

The correct option is B 1.164 g
125 mL of 0.04 M AgNO3=125×0.04 millimoles
=125×0.041000 mol
Molar mass of AgNO3=170 g/mol
weight of pure AgNO3=125×0.04×1701000 g

It is given in the question that AgNO3 is 73% by mass which means 100 g AgNO3 sample contains 73 g pure AgNO3

Total weight of sample required=125×0.04×170×1001000×73 g
=1.164 g

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