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Standard IX

Mathematics

Application of Direct Proportionality

AgNO3 sample ...

Question

$A g N O_{3}$ sample is $73 %$ by mass. To prepare $125 m L$ of $0.04 M A g N O_{3}$ solution, $A g N O_{3}$ sample required is

1.471 g

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1.164 g

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1.250 g

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1.063 g

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Solution

The correct option is B $1.164 g$
$125 m L of 0.04 M A g N O_{3} = 125 \times 0.04 millimoles$
$= \frac{125 \times 0.04}{1000} m o l$
$Molar mass of A g N O_{3} = 170 g / m o l$
$weight of pure A g N O_{3} = \frac{125 \times 0.04 \times 170}{1000} g$

It is given in the question that $A g N O_{3}$ is $73 %$ by mass which means $100 g A g N O_{3}$ sample contains $73 g$ pure $A g N O_{3}$

$∴ Total weight of sample required = \frac{125 \times 0.04 \times 170 \times 100}{1000 \times 73} g$
$= 1.164 g$

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Similar questions

A g N O_{3}

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73 %

by mass. To prepare

125 m L

0.04 M A g N O_{3}

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A g N O_{3}

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A g N O_{3}

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K O C l

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A g N O_{3}

is dissolved in

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50

mL of

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precipitated is filtered off. Excess of

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K I O_{3}

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mL of the same stock solution of

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30

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A g N O_{3}

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K I O_{3} + 2 K I + 6 H C l ⟶ 3 I C l + 3 K C l + 3 H_{2} O

Q. The amount of

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Application of Direct Proportionality

Standard IX Mathematics

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

$A g N O_{3}$ sample is $73 %$ by mass. To prepare $125 m L$ of $0.04 M A g N O_{3}$ solution, $A g N O_{3}$ sample required is