The concentration of 1 L of CaCO_3 solution | vedclass

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(D) $ppm$ (parts per million) $= \frac{	ext{mass of solute in mg}}{	ext{volume of solution in L}}$Given,$ppm = 1000$Volume $= 1 \ L$Mass of solute $

Vedclass · Accepted Answer

$10^{-2}$

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(D) $ppm$ (parts per million) $= \frac{	ext{mass of solute in mg}}{	ext{volume of solution in L}}$Given,$ppm = 1000$Volume $= 1 \ L$Mass of solute $= 1000 \ mg = 1 \ g$Molar mass of $CaCO_3 = 40 + 12 + (3 	imes 16) = 100 \ g \ mol^{-1}$Number of moles $= \frac{	ext{mass}}{	ext{molar mass}} = \frac{1 \ g}{100 \ g \ mol^{-1}} = 0.01 \ mol = 10^{-2} \ mol$Concentration $= \frac{	ext{moles}}{	ext{volume in L}} = \frac{10^{-2} \ mol}{1 \ L} = 10^{-2} \ mol \ L^{-1}$

The concentration of $1 \ L$ of $CaCO_3$ solution is $1000 \ ppm$. What is its concentration in $mol \ L^{-1}$ ?
$(Ca=40 \ u, O=16 \ u, C=12 \ u)$

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The concentration of $1 \ L$ of $CaCO_3$ solution is $1000 \ ppm$. What is its concentration in $mol \ L^{-1}$ ?$(Ca=40 \ u, O=16 \ u, C=12 \ u)$