Some Basic Concepts of Chemistry 2 Question 5
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5. $100 \mathrm{~mL}$ of a water sample contains $0.81 \mathrm{~g}$ of calcium bicarbonate and $0.73 \mathrm{~g}$ of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of $\mathrm{CaCO}_{3}$ is (molar mass of calcium bicarbonate is $162 \mathrm{~g} \mathrm{~mol}^{-1}$ and magnesium bicarbonate is $146 \mathrm{~g} \mathrm{~mol}^{-1}$ )
======= ####5. $100 \mathrm{~mL}$ of a water sample contains $0.81 \mathrm{~g}$ of calcium bicarbonate and $0.73 \mathrm{~g}$ of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of $\mathrm{CaCO}_{3}$ is (molar mass of calcium bicarbonate is $162 \mathrm{~g} \mathrm{~mol}^{-1}$ and magnesium bicarbonate is $146 \mathrm{~g} \mathrm{~mol}^{-1}$ )
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $5,000 \mathrm{ppm}$
(b) $1,000 \mathrm{ppm}$
(c) $100 \mathrm{ppm}$
(d) $10,000 \mathrm{ppm}$
(2019 Main, 8 April I)
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Solution:
- Given, $W_{\mathrm{Ca}\left(\mathrm{HCO}{3}\right){2}}=0.81 \mathrm{~g}$
$$ \begin{aligned} W_{\mathrm{Mg}\left(\mathrm{HCO}{3}\right){2}} & =0.73 \mathrm{~g} \ M_{\mathrm{Ca}\left(\mathrm{HCO}{3}\right){2}} & =162 \mathrm{~g} \mathrm{~mol}^{-1}, \ M_{\mathrm{Mg}\left(\mathrm{HCO}{3}\right){2}} & =146 \mathrm{~mol}^{-1} \end{aligned} $$
$$ \begin{aligned} V_{\mathrm{H}{2} \mathrm{O}} & =100 \mathrm{~mL} \ \text { Now, } \quad n{\mathrm{eq}}\left(\mathrm{CaCO}{3}\right) & =n{\mathrm{eq}}\left[\mathrm{Ca}\left(\mathrm{HCO}{3}\right){2}\right]+n_{\mathrm{eq}}\left[\mathrm{Mg}\left(\mathrm{HCO}{3}\right){2}\right] \ \frac{W}{100} \times 2 & =\frac{0.81}{162} \times 2+\frac{0.73}{146} \times 2 \ \therefore \quad \frac{W}{100} & =0.005+0.005 \ W & =0.01 \times 100=1 \end{aligned} $$
Thus, hardness of water sample $=\frac{1}{100} \times 10^{6}=10,000 \mathrm{ppm}$
