Question

Hardness due to 13.6 gm/lit of CaSO4 can be expressed in terms of CaCO3
is
O 20 mg/lit
O 10 mg/lit
O 2 mg/lit
5 mg/lít

Answer 1

Answer:

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Answer 2

Answer:

The hardness due to $13.6mg/L$ of $CaSO_{4}$ in terms of $CaCO_{3}$ = $13.6\times \frac{100}{136}$ = $10mg/L$.

Therefore, option b) $10mg/L$ is correct.

Explanation:

Given,

The weight of $CaSO_{4}$ = $13.6mg/L$

The hardness due to $13.6mg/L$ of $CaSO_{4}$ in terms of $CaCO_{3}$ =?

As we know,

• The hardness caused due to the sulfates of calcium and magnesium is a permanent hardness.

Now, to express it in terms of $CaCO_{3}$, we have to multiply it with the formula given below:

• $CaCO_{3}$ equivalent = $\frac{M_{CaCO_{3}} }{M_{CaSO_{4}} }$

Here, $M_{CaCO_{3}}$ is the molar mass of $CaCO_{3}$ = $100g/mol$

And $M_{CaSO_{4}}$ is the molar mass of $CaSO_{4}$ = $136g/mol$

Therefore,

• $CaCO_{3}$ equivalent = $\frac{M_{CaCO_{3}} }{M_{CaSO_{4}} }$ = $\frac{100}{136}$

And,

The hardness due to $13.6mg/L$ of $CaSO_{4}$ in terms of $CaCO_{3}$ = $13.6\times \frac{100}{136}$ = $10mg/L$.