a. `N_(1) V_(1) + N_(2) V_(2) + N_(3) V_(3) = N_(4) V_(4)`
`(V_(4) = V_(1) + V_(2) V_(3) = 100 + 100 + 100 = 300 mL)`
`n` factor for `HCl` and `HNO_(3) = 1`
`n` factor for `H_(2) SO_(4) = 2`
In such questions always convert molarity to normality, since 1 grams equivalent of one reactant is always equal to 1 gram equivalent of another reactant.
`100 xx 0.1 + 100 xx 0.2 + 2 100 xx 0.1 = N_(4) xx 300 + 10 + 40 + 10 = N_(4) xx 300`
`:. N_(4) = (60)/(300) = (1)/(5) = 0.2 N`
Hence, final concentration of solution `= 0.2 N`
b. Final volume `= 1 L = 1000 mL`
`:. 100 xx 0.1 + 100 xx 0.2 xx 2 + 100 xx 0.1 = N_(4) xx 1000`
`N_(4) = (60)/(1000) = 0.06 N`
Hence, final concentration of solution `= 0.06 N`
c. Final volume `= 300 + 700 = 1000 mL`
`:. N_(4) = (60)/(1000) = 0.06`
Hence, final concentration of solution `= 0.06 N`