Question

find equivalent weight of metal when metal carbonate is neutralized by 10ml of 0.1N HCl and further the resulting metal chloride gave 0.0519gm of metal phosphate.

Answer 1

Answer:

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.