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Question

Chemistry Question on Some basic concepts of chemistry

One litre hard water contains 12.00 mg Mg2+Mg^{2+}. Milliequivalents of washing soda required to remove its hardness is

A

1

B

12.16

C

1×10−31 \times 10^{-3}

D

12.16×10−312.16 \times 10^{-3}

Answer

1

Explanation

Solution

1g−eqMG2++   1g−eqNa2CO3→MgCo3+2Na+_{1g - eq}^{MG^{2+}} + _{\, \, \, 1g - eq}^{Na_2CO_3} \rightarrow MgCo_3 + 2Na^+ 1 g-equivalent of Mg2+=12 g of Mg2+Mg^{2+} = 12 \, g \, of \, Mg^{2+} =12000 mg of Mg2+= 12000 \, mg \, of \, Mg^{2+} Now, 12000 mg of Mg2+≡100Mg^{2+} \equiv 100 milliequivalent of            Na2CO3 \, \, \, \, \, \, \, \, \, \, \, Na_2 CO_3 ∴12mg of Mg2+≡milliequivalent of Na2Co3\therefore 12 mg \, of \, Mg^{2+} \equiv milliequivalent \, of \, Na_2Co_3