If 200 ml of `0.1` M `Na_(2)SO_(4)` is mixed with 100 ml of `0.2` M `NA_(3)PO_(4)` solution, molarity of `Na^(+)` in the final solution, if final solution has density `1.2gm//ml`, will be :

Options

A. `0.196 M`
B. `0.33 M`
C. `0.5 M`
D. None of these

Answer

B

Explanation

Moles of `Na_(2)SO_(4)=200xx0.1xx10^(-3)`
Moles of `Na^(+)=(200xx0.1xx10^(-3))xx2=(2xx10^(-2))xx2`
Moles of `Na_(3)PO_(4)=100xx0.2xx10^(-3)`
Moles of `Na^(+)=(100xx0.2xx10^(-3))xx3=2xx10^(-2)xx3`
Total moles of `Na^(+)=5xx2xx10^(-2)=10^(-1)`
`M_(Na^(+))=(10^(-1))/(300)xx1000=1/3=0.33M`