One litre of sample of hard water contains 1 mg of `CaCl_(2)` and 1 mg of `MgCl_(2).` Find the total hardness of water in terms of `CaCO _(3)` present per `10^6` parts of water by mass.

Options

A. 2.5ppm
B. 1.95ppm
C. 2.15ppm
D. 195ppm

Answer

2

Explanation

mM of `M g Cl _(2) = ( 1 xx 10 ^(-3) xx 10 ^(3))/( 95) ` (milli mole `= ("mass")/("Molar mass") xx 1000)`
`m M of Ca Cl _(2) = (1 xx 10 ^(-3) xx 10 ^(3))/( 111) = (1)/(111)`
`therefore m M of Ca CO _(3) if Mg Cl _(2) and Ca Cl _(2) ` are taken in form of `C a CO _(3) = (1)/( 95) + (1)/(111) = ( 1 0 6)/( 111 xx 95)`
`(because Ca, `Mgare both bivalent `therefore` mole ratio is `1:1), therefore `Mass of `Ca CO _(3)` in `100 mL = ( 206)/(111xx 95) xx (100)/(1000)`
`therefore ` Hardness in pp ( part per million) `= ( g of Ca CO _(3))/( 10 ^(6) g of H _(2) O ) = ( 206 xx 100 xx 10 ^(6))/( 111 xx 95 xx 1000 xx 1000) = 1. 953 = 2 ` ppm