`100 mL` of a sample of hard water requires `25.1 mL` of `0.02 N H_(2)SO_(4)` for complete reaction, The hardness of water ( density `1g//mL)` is:

Options

A. 200 ppm
B. 250 ppm
C. 251 ppm
D. 258 ppm

Answer

C

Explanation

milliequivalent of `H_(2)SO_(4)=25.1xx0.02=0.502`
Equivalent `=0.502xx10^(-3)`
Eq of `H_(2)SO_(4)=`Eq of `CaCO_(3)`
`therefore` Mass of `CaCO_(3)=Eqxx50=0.502xx10^(-3)xx50=0.0251`
100 mL of `H_(2)O=100g` contains 0.0251 g `CaCO_(3)`
`10^(6)g` of water will contain `(0.0251)/(10^(2))xx10^(6)=251g` of `CaCO_(3)`
`therefore` The degree of hardness `=251`ppm