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10.875 g of a mixture of NaCl and Na(2)C...

10.875 g of a mixture of NaCl and `Na_(2)CO_(3)` was dissolved in water and the volume made up to 250 mL, 20 mL of this solution required 75.5 mL of `(N)/(10) H_(2)SO_(4)`. Find out the percentage composition of the mixture.

Text Solution

Verified by Experts

Only `Na_(2)CO_(3)` will react `H_(2)SO_(4)`.
Applying `underset((Na_(2)CO_(3)))(N_(1)V_(1)-=underset((H_(2)SO_(4)))(N_(2)V_(2))`
`N_(1)xx20=75.5xx(1)/(10)`
`N_(1)=(75.5)/(20xx10)=0.3775`
`underset(1 " mol.mass")(Na_(2)CO_(3))+underset(2g eq.)(H_(2)SO_(4)) to Na_(2)SO_(4)+H_(2)O+CO_(2)`
Eq. mass of `Na_(2)CO_(3)=(106)/(2)=53`
Mass of `Na_(2)CO_(3)` present in 250 mL 0.3775 N solution
`=(NxxExxV)/(1000)=(0.3775xx53xx250)/(1000)`
=5.0018 g
Mass of NaCl`=(10.875-5.0018)=5.8732 g`
`Na_(2)CO_(3)=(5.0018)/(10.875)xx100=45.99%`
`NaCl=(5.8732)/(10.875)xx100=54.0%`
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