Calculate the degree of hardeners of river water whose `100 mL` solution required `1.68 mL` of `0.1 N H_(2)SO_(4)`.

Explanation

Meq. Of `H_(2)SO_(4)` used `= 1.68 xx 0.1 = 0.168`
Assuming `CaCO_(3)` in water, Meq. Of `CaCO_(3) = 0.168`
`:.(w)/(50) xx 1000 = 0.168`
`:. w_(CaCO_(3)) = 8.4 xx 10^(-3) g`
`because 100 mL` water `-= 8.4 xx 10^(-3)g CaCO_(3)`
`:. 10^(6) mL` water `-= (8.4 xx 10^(-3) xx 10^(6))/(100)`
`= 8.4 xx 10 = 84`ppm