Question

What is the degree of hardness in ppm of a sample containing \(19\) mg of \(MgCl_2\) \((\text{Molecular Weight}=95)\) in \(2\) kg water sample? Express it in terms of equivalents of \(CaCO_3\).

• A. \(10\)
• B. \(20\)
• C. \(30\)
• D. \(40\)

Hint:

Hardness is commonly expressed as \(CaCO_3\) equivalent in ppm.

Answer 1

The Correct Option is A

Step 1: Molecular weight of \(MgCl_2\) is: \[ 95 \]

Step 2: Mass of \(MgCl_2\) present: \[ 19\,mg \]

Step 3: Equivalent weight of \(MgCl_2\): \[ \frac{95}{2}=47.5 \]

Step 4: Equivalent weight of \(CaCO_3\): \[ \frac{100}{2}=50 \]

Step 5: Convert \(MgCl_2\) hardness into \(CaCO_3\) equivalent: \[ \text{Mass as }CaCO_3= 19\times \frac{50}{47.5} \] \[ =20\,mg \]

Step 6: Water sample is \(2\) kg, which is approximately \(2\) litres.

Step 7: Hardness in ppm: \[ \text{ppm}=\frac{20}{2} \] \[ =10 \] \[ \boxed{10} \]