A water sample contains, per litre, 48.6 mg Ca(HCO3)2, 36.5 mg Mg(HCO3)2 and 20.4 mg MgSO4. Hardness is expressed as ppm CaCO3 (mg of CaCO3 per litre containing the same number of moles of divalent cations; CaCO3=100). Use molar masses Ca(HCO3)2=162, Mg(HCO3)2=146, MgSO4=120 g mol^-1. The total hardness is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3eq017

A water sample contains, per litre, $48.6 mg$ $Ca (HCO X_{3}) X_{2}$ , $36.5 mg$ $Mg (HCO X_{3}) X_{2}$ and $20.4 mg$ $MgSO X_{4}$ . Hardness is expressed as ppm $CaCO X_{3}$ (mg of $CaCO X_{3}$ per litre containing the same number of moles of divalent cations; $CaCO X_{3} = 100$ ). Use molar masses $Ca (HCO X_{3}) X_{2} = 162$ , $Mg (HCO X_{3}) X_{2} = 146$ , $MgSO X_{4} = 120 g mo l^{- 1}$ . The total hardness is:

Each salt has one divalent cation per formula unit, so each mg is converted to mg

CaCO X_{3}

\times 100/ M

48.6 \times 100/162 = 30.0

36.5 \times 100/146 = 25.0

20.4 \times 100/120 = 17.0

. Total

= 72.0

ppm as

CaCO X_{3}

(simply adding the masses,

105.5

, ignores the conversion).