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0.50 g of a mixture of K(2)CO(3) and Li(...

`0.50 g` of a mixture of `K_(2)CO_(3)` and `Li_(2)CO_(3)` required `30 mL` of `0.25 N HCl` solution for neutralization. What is `%` composition of mixure?

Text Solution

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`underset(138)(K_(2)CO_(3))+underset(2xx36.5)(2HCl) to 2KCl+H_(2)O+CO_(2)`
Eq. mass 69 `" "`36.5
`underset(74)(Li_(2)CO_(3))+underset(2xx36.5)(2HCl) to 2LiCl+H_(2)O+CO_(2)`
Eq. mass 37 `" " `36.5
Let x g of `K_(2)CO_(3)` be present in the mixture.
Mass of `Li_(2)CO_(3)=(0.50-x)`
No. of g equivalents of `K_(2)CO_(3)=(x)/(69)`
No. of g equivalents of `Li_(2)CO_(3)=((0.50-x))/(37)`
No. of g equivalents in 30 mL of 0.25 N HCl
`("Normality"xx"Volume")/(1000)=(0.25xx30)/(1000)=(3)/(400)`
At equivalence point,
`(x)/(69)+((0.50-x))/(3)/(400)`
So, x=0.48
`K_(2)CO_(3)=0.48 g `, or 96%
`Li_(2)CO_(3)=0.02 g`, or 4 %
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