Question

MgCl$_2$ and CaSO$_4$ salts are added to 1 litre of distilled deionized water and mixed until completely dissolved. Total Dissolved Solids (TDS) concentration is 500 mg/l, and Total Hardness (TH) is 400 mg/l (as CaCO$_3$). The amounts of MgCl$_2$ and CaSO$_4$ added are calculated (rounded off to the nearest integer). Which of the following options is/are true:

• A. Amount of MgCl$_2$ added is 143 mg
• B. Amount of CaSO$_4$ added is 357 mg
• C. Amount of MgCl$_2$ added is 103 mg
• D. Amount of CaSO$_4$ added is 397 mg

Hint:

In such problems, make sure to carefully solve for the unknowns using the system of equations derived from the molar ratios, and then apply the results to calculate the required amounts.

Answer 1

The Correct Option is C, D

Given Data:

• Total Dissolved Solids (TDS) = 500 mg/l
• Total Hardness (TH) = 400 mg/l (as CaCO₃)
• Salts added: MgCl₂ and CaSO₄

Step 1: Dissociation Equations

\[ \text{MgCl}_2 \rightarrow \text{Mg}^{2+} + 2\text{Cl}^- \]

Molar mass of MgCl₂ = 95 g/mol

Molar mass of CaSO₄ = 136 g/mol

Step 2: Total Hardness Calculation

The formula for total hardness (TH) as CaCO₃ is:

\[ \text{TH} = \left( \frac{\text{Ca}^{2+}}{20} + \frac{\text{Mg}^{2+}}{12} \right) \times 50 \]

From the given data:

\[ 400 = \left( \frac{40B}{20} + \frac{24A}{12} \right) \times 50 \]

Simplifying:

\[ 2A + 2B = 8 \quad \text{and} \quad A + B = 4. \]

Step 3: Solving for A and B

Solving the equations:

\[ A = 1.073, \quad B = 2.926. \]

Step 4: Calculating the Amount of MgCl₂ and CaSO₄

\[ \text{MgCl}_2 = 95 \times A = 95 \times 1.073 = 103 \text{ mg/l} \]

\[ \text{CaSO}_4 = 136 \times B = 136 \times 2.926 = 397 \text{ mg/l} \]

Thus, the correct answers are (C) and (D).