A water sample contains 30 , ppm , MgSO_4 and | vedclass

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(A) The molar mass of $MgSO_4$ is $120 \, g/mol$. Since $120 \, g \, MgSO_4$ is equivalent to $100 \, g \, CaCO_3$,the $30 \, ppm \, MgSO_4$ is equiva

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(A) The molar mass of $MgSO_4$ is $120 \, g/mol$. Since $120 \, g \, MgSO_4$ is equivalent to $100 \, g \, CaCO_3$,the $30 \, ppm \, MgSO_4$ is equivalent to $\frac{100}{120} 	imes 30 = 25 \, ppm \, CaCO_3$. The molar mass of $MgCl_2$ is $95 \, g/mol$. Since $95 \, g \, MgCl_2$ is equivalent to $100 \, g \, CaCO_3$,the $19 \, ppm \, MgCl_2$ is equivalent to $\frac{100}{95} 	imes 19 = 20 \, ppm \, CaCO_3$. Total hardness of water = $25 + 20 = 45 \, ppm \, CaCO_3$ equivalent.

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

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