Question

Solution of 100 ml water contains 0.73 g of Mg(HCO₃)₂ and 0.81 g of Ca(HCO₃)₂. Calculate the hardness in terms of ppm of CaCO₃.

• A. 10² ppm
• B. 10³ ppm
• C. 10⁴ ppm
• D. 5 × 10³ ppm

Answer 1

The correct option is C

10⁴ ppm

Explanation for correct answer:-

The correct answer is option C.

Step 1:- To calculate the total no. of moles present in the given solution.

$\mathrm{ppm}(\mathrm{Parts}\mathrm{Per}\mathrm{Million})=\frac{\mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{moles}\times 100\times {10}^6}{100}$

Molar mass = The sum of the total mass in grams of the atoms present to make up a molecule per mole.

$$\begin{gathered} \mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{Mg}{\left({\mathrm{HCO}}_3\right)}_2=24.3+1\times 2+12\times 2+16\times 6=146{{\mathrm{gmol}}^{-1}}^{} \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2=40+1\times 2+12\times 2+16\times 6=162{{\mathrm{gmol}}^{-1}}^{} \end{gathered}$$

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{Mg}{\left({\mathrm{HCO}}_3\right)}_2=\frac{\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{substance}}{\mathrm{Molar}\mathrm{mass}} \\ =\frac{0.73}{146} \\ =0.005\mathrm{moles} \end{gathered}$$

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2=\frac{\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{substance}}{\mathrm{Molar}\mathrm{mass}} \\ =\frac{0.81}{162} \\ =0.005\mathrm{moles} \end{gathered}$$

$\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{in}\mathrm{a}100\mathrm{ml}\mathrm{water}=0.005+0.005=0.01\mathrm{moles}$

Step 2 :- Calculation of hardness in ppm of ${\mathrm{CaCO}}_3$

$$\begin{gathered} \mathrm{ppm}=\frac{\mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{moles}\times 100\times {10}^6}{100} \\ =\frac{0.01\times 100\times {10}^6}{100} \\ ={10}^4\mathrm{ppm} \end{gathered}$$