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

Question

One litre hard water contains 12.00 mg Mg2+. Milliequivalents of washing soda required to remove its hardness is (A) 1 (B) 12.16 (C) 1 × 10-3 (D) 12.16 × 10-3

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/d92d221d6ccb892a3f996ef389144535-.png" /> Meq of Mg +2= Meq of washing soda ( W / E ) × 1000= Mg +2 ; EW =(24/2)=12 (12 × 10-3/12) × 1000=1

Q. One litre hard water contains $12.00 m g M g^{2 +}$ . Milliequivalents of washing soda required to remove its hardness is

$1$

$12.16$

$1 \times 1 0^{- 3}$

$12.16 \times 1 0^{- 3}$

Meq of $M g^{+ 2} =$ Meq of washing soda
$\frac{W}{E} \times 1000 = M g^{+ 2}; E W = \frac{24}{2} = 12$
$\frac{12 \times 1 0 ^{- 3}}{12} \times 1000 = 1$

Q. One litre hard water contains 12.00mgMg2+. Milliequivalents of washing soda required to remove its hardness is

1

12.16

1×10−3

12.16×10−3

Meq of Mg+2= Meq of washing soda EW​×1000=Mg+2;EW=224​=12 1212×10−3​×1000=1

Q. One litre hard water contains $12.00 m g M g^{2 +}$ . Milliequivalents of washing soda required to remove its hardness is

$1$

$12.16$

$1 \times 1 0^{- 3}$

$12.16 \times 1 0^{- 3}$

Meq of $M g^{+ 2} =$ Meq of washing soda
$\frac{W}{E} \times 1000 = M g^{+ 2}; E W = \frac{24}{2} = 12$
$\frac{12 \times 1 0 ^{- 3}}{12} \times 1000 = 1$