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0.9031 g of a mixture of NaCI and KCI on...

0.9031 g of a mixture of NaCI and KCI on treatment with `H_2SO_4` yielded 1.0784 g of a mixture of `Na_2SO_4` and `K_2SO_4`.Calculate the percent composition of the mixture.

Text Solution

Verified by Experts

Let the amount of NaCl in the mixture be x g and KCl be 0.9031 - x g.
We know.
`2underset(117)(NaCl)+H_(2)SO_(4)rarr underset(142)(Na_(2)SO_(4))+2HCl`
`2underset(149)(KCl)+H_(2)SO_(4)rarr underset(174)(K_(2)SO_(4))+2HCl`
x g of NaCl will give `(142)/(117)` x g of `Na_(2)SO_(4)`
(0.9031 - x) g of KCl will give `(174)/(149)(0.9031 - x)g` of `K_(2)SO_(4)`
Thus, `(142)/(117)x +(174)/(149)(0.9031-x)=1.0784`
`rArr x = 0.518 g`
Now, % of `NaCl = (0.518)/(0.9031)xx100`
`= 57.36%`
and % of KCl = 42.64 % .
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