A sample of water contains 30 ppm of MgSO_4 | vedclass

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(A) Hardness is expressed in terms of $CaCO_3$ equivalents.Formula: $	ext{Hardness} = 	ext{Mass of salt} 	imes \frac{	ext{Molar mass of } CaCO_3}{

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$45$

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(A) Hardness is expressed in terms of $CaCO_3$ equivalents.Formula: $	ext{Hardness} = 	ext{Mass of salt} 	imes \frac{	ext{Molar mass of } CaCO_3}{	ext{Molar mass of salt}}$.$1$. For $MgSO_4$ $(M = 120 \ g/mol)$: $	ext{Hardness} = 30 	imes \frac{100}{120} = 25 \ ppm$.$2$. For $MgCl_2$ $(M = 95 \ g/mol)$: $	ext{Hardness} = 19 	imes \frac{100}{95} = 20 \ ppm$.Total hardness = $25 + 20 = 45 \ ppm$.

$A$ sample of water contains $30 \ ppm$ of $MgSO_4$ and $19 \ ppm$ of $MgCl_2$. Calculate the hardness of water in $ppm$ (in terms of $CaCO_3$ equivalents).

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