0.5 g of a mixture of `K_2CO_3` and `Li_2CO_3` requires 30 " mL of " 0.25 N Hci for neutralisaion. Calculate the percentage composition of the mixture.

Options

A. `96%K_(2)CO_(3),4%Li_(2)CO_(3)`
B. `4%K_(2)CO_(4),96%Li_(2)CO_(3)`
C. `48%K_(2)CO_(3),52%Li_(2)CO_(3)`
D. `52%K_(2)CO_(3),48%Li_(2)CO_(3)`

Answer

A

Explanation

Mw of `K_(2)CO_(3)=39xx2+12+48=138 g mol^(-1)`
`mE of K_(2)CO_(3)=(138)/(2)=69g`
Mw of `LiC_(2)CO_(3)=7xx2+12+48=74 g mol^(-1)`
Ew of `Li_(2)CO_(3)=(74)/(2)=37g`
Let x g of `K_(2)CO_(3)` and `(0.5-x) g of Li_(2)CO_(3)`
`mEq K_(2)CO_(3)+m" Eq of "Li_(2)CO_(3)=m" Eq of "HCl`
`((x)/(69)+(0.5-x)/(37))xx1000=30xx0.25`
`(37x+69(0.5-x))/(69xx37)=(30xx0.25)/(1000)`
`69xx0.5-(30xx0.25xx69xx37)/(1000)=32x`
`34.5-19.14=32x`
`32x=15.36 and x=0.48`
`% of K_(2)CO_(3)=(0.48xx100)/(0.5)=96%`
`% of Li_(2)CO_(3)=4%`