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  • If 1 L hard water contains 18.00 mg . Find the milli equivalent of was

If 1 L hard water contains 18.00 mg \mathrm{Mg}^{2+}. Find the milli equivalent of washing soda to remove the hardness.

Option: 1

1.0


Option: 2

1.5


Option: 3

2.0


Option: 4

2.5


Answers (1)

best_answer

\mathrm{M_{\text {eq }} \text { of } \mathrm{Mg}^{2+}=M_{\text {eq }} \text { of waining soda }}

\mathrm{\frac{W}{E} \times 1000=\mathrm{Mg}^{2+} ; \quad E W=\frac{24}{2}=12}

\mathrm{\therefore \quad \frac{18 \times 10^{-3}}{12} \times 1000=1.5}

 

Posted by

Pankaj Sanodiya

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