`5 mL` of `8 N HNO_(3), 4.8 mL` of `5N HCl` and a certain volume of `17 M H_(2)SO_(4)` are mixed together and made upto `2 "litre"`. `30 mL` of this acid mixture exactly neutralizes `42.9 mL` of `Na_(2)CO_(3)` solution containing `1 g Na_(2)CO_(3)`. `10 H_(2)O "in" 100 mL` of water. Calculate the amount of sulphate ions in `g` present in solution.

`5 mL` of `8 N HNO_(3), 4.8 mL` of `5N HCl` and a certain volume of `1

1.	`5 mL` of `8 N HNO_(3), 4.8 mL` of `5N HCl` and a certain volume of `17 M H_(2)SO_(4)` are mixed together and made upto `2 "litre"`. `30 mL` of this acid mixture exactly neutralizes `42.9 mL` of `Na_(2)CO_(3)` solution containing `1 g Na_(2)CO_(3)`. `10 H_(2)O "in" 100 mL` of water. Calculate the amount of sulphate ions in `g` present in solution.
Answer» For `H_(2)SO_(4)`, n factor`=2` Let there be `n m" Eq of "H_(2)SO_(4)`. m" Eq of "all acids in `2L=(5xx8)+(48.5xx5)+n` `=64+n` Normality`=((6n+n))/(2000)` Normality of `Na_(2)CO_(3).10H_(2)O` is `(1)/((286)/(2))xx(1)/(100)xx1000=0.07` `thereforeN_(1)V_(1)=N_(2)V_(2)` or `((64+n))/(200)xx30=0.07xx42.9` or `n=136.2 m" Eq of "H_(2)SO_(4)` Moles of `H_(2)SO_(4)=(136.2)/(2)xx10^(-3)` Mass of `SO_(4)^(2-)=(136.2)/(2)xx10^(-3)xx96` `=6.5376g`

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The purity of `H_2O_2` in a given sample is `85%.` Calculate the weight of impure sample of `H_2O_2` which requires `10mL` of `M//5 KMnO_4` solution in a titration in acidic mediumA. 2 gB. 0.2 gC. 0.17 gD. 0.15 g
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