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Common Ion Effect

Weight of N...

Question

Weight of ${N a}_{2} {C O}_{3}$ which can be neutralised by 100 ml of 0.05M $H_{2} {S O}_{4}$ is:

A

0.53 g

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B

1.06 g

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C

5.3 g

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D

10.6 g

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Solution

The correct option is A $0.53 g$
Using stoichiometry,

$E q_{H_{2} S O_{4}} = E q_{N a_{2} C O_{3}}$

$2 \times 0.05 \times \frac{100}{1000} = 2 \times m o l e s_{(} N a_{2} C O_{3})$

$M o l e s o f N a_{2} C O_{3} = \frac{5}{1000}$

Therefore, weight of $N a_{2} C O_{3} = 106 \times \frac{5}{1000} g m = 0.53 g m$

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Similar questions

100 m l

{N a}_{2} {C O}_{3}

solution containing

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l o g 5 = 0.6990

1.25 g

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{N a}_{2} {S O}_{4}

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25 m l

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Q. A solution contain

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N a H C O_{3}

. 10 ml of this solution required 2.5 ml of 0.1 M

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{N a}_{2} {S O}_{4}

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25 m l

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{N a}_{2} {C O}_{3} = 106

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