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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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Let the amount of` K_(2)CO_(3)` be 'x' g
`therefore "Amount of" Li_(2)CO_(3)=(0.5-x)g`
`"Number of equivalents"=(x)/(138//2)+((0.5-x))/(74//2)....(i)`
a Number of equivalents of HCl used `=(NV)/(1000)=(0.25xx30)/(1000)=7.5xx10^(-3)...(ii)`
Comparing eqs. (i) and (ii) we get x=0.48g
`therefore "Mass of" K_(2)Cr_(3)=0.48g`
Mass of `LiCO_(3)=0.02g`
`% K_(2)CO_(3)=(0.48)/(0.5)xx100=96`
`% Li_(2)CO_(3)=(0.02)/(0.5)xx100=4`
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