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0.5 g of a mixture of K2CO3 and Li2CO3 r...

0.5 g of a mixture of `K_2CO_3` and `Li_2CO_3` requires 15 " mL of " 0.25 N Hci for neutralisaion. Calculate the percentage composition of the mixture.

Text Solution

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" Eq of "HCl `=(0.25xx30)/(1000)`
Let x g of `K_2CO_3 and (0.5-x)g of Li_2CO_3` reacts.
" Eq of "`K_2CO_3=(x)/(69), " Eq of "Li_2CO_3=(0.5-x)/(37)`
`therefore(x)/(60)+(0.5-x)/(37)=(0.25xx30)/(1000)impliesx=0=0.48g`
`% of K_2CO_3=(0.48)/(05)xx100=96%`
`% of Li_2CO_3=4%`
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