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1.0 g of the carbonate of a metal was dissolved in 25 ml of 1.0 n HCl. The resulting liquid required 5 ml of 1.0 n NaOH for neutralization. The equivalent weight of the metal carbonate is: [2.5 M]rn(A) 50rn(b) 30rn(c)rn20rn(d)rnnone of these

Accepted Answer
The given question and options are given below:

1.0 g of the carbonate of a metal was dissolved in 25 mL of 1.0 N HCl. The resulting liquid required 5 mL of 1.0 N NaOH for neutralization. The equivalent weight of the metal carbonate is: [2.5 M]
(A) 50
(B) 30
C) 20
(D) None of these
The correct answer is (C) 20.

Explanation:

Equivalent weight of metal carbonate = (Weight of metal carbonate / Equivalents of metal carbonate)

Equivalents of metal carbonate = Equivalents of HCl used - Equivalents of NaOH used

Equivalents of HCl used = (25 mL * 1.0 N) = 25 meq

Equivalents of NaOH used = (5 mL * 1.0 N) = 5 meq

Therefore, Equivalents of metal carbonate = 25 meq - 5 meq = 20 meq

Weight of metal carbonate = 1.0 g

Therefore, Equivalent weight of metal carbonate = (1.0 g / 20 meq) = 20

Hence, the equivalent weight of the metal carbonate is 20.

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