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1g of the carbonate of a metal was dissolved in 25 mL of N-HCl. The resulting liquid required 5 mL of N-NaOH for neutralisation. The equivalent weight of the metal carbonate is 50 30 20 None

1g of the carbonate of a metal was dissolved in 25 mL of N-HCl. The resulting liquid required 5 mL of N-NaOH for neutralisation. The equivalent weight of the metal carbonate is
  1. 50
  2. 30
  3. 20
  4. None

Grade:11

1 Answers

Arun
25763 Points
3 years ago
For 5ml of 1 N NaOH, HCl required is:
M1V1 = M2V2
1*5 = 1*V2
V2 = 5 ml
So, HCl required will be 5 ml.
So, volume of HCl left = 25-5 = 20 ml = 0.020 L
Concentration of HCl = 1 N
For HCl, 1N = 1 M, because n-factor = 1
molarity = number of moles/volume
1 M = ​number of moles/0.020
number of moles = 0.020 mol
0.020 mol of HCl will be used in neutralization of carbonate.
The reaction is:
M2CO3 + 2HCl ----> 2MCl + H2CO3
1 mol of M2CO3 is neutralized by 2 moles of HCl. If 0.020 mol of HCl is used, then M2CO3 used = 0.010 mol
mass of M2CO3 = 1 g
moles = 0.010 mol
molar mass = mass/moles = 1/0.010 = 100 g
Molar mass = 100 g/mol
n-factor = 2 (for carbonate)
equivalent weight = 100/2 = 50
Equivalent weight of metal carbonate = 50
correct option is 1.
 

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