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2.5 g of the carbonate of a metal was tr...

2.5 g of the carbonate of a metal was treated with 100 mL of 1 N `H_2 SO_4`. After the completion of the reaction, the solution was boiled off to expel `CO_2` and was then titrated against 1 N NaOH solution. Calculate the volume of alkali that would be consumed, if the equivalent weight of the metal is 20.

A

50

B

25

C

75

D

100

Text Solution

Verified by Experts

The correct Answer is:
A
Equivalent weight of metal carbonate = 20 + 30 =50 2.5 g of metal carbonate `=(2.5 )/(50 ) = 0.05 `eq
Number of equivalent of `H_2 SO_4` that would have reacted = 0.05
Number of equivalent of `H_2 SO_4` taken `=(100 xx 1)/(1000 ) = 0.1`
Number of equivalent of `H_2 SO_4` which remains unreacted = 0.1 -0.05 = 0.05 eq.
` therefore ` Number of equivalent of alkali consumed = 0.05 eq, millieq. = Normality ` xx` Volume in mL
` therefore 1.0 xx V = 0.05 xx 1000 V= ( 0.05 xx 1000 )/( 1.0 )= 50 mL `
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