For 50ml of 1 N/10 NaOH, HCl required is:
M1V1 = M2V2
1/10*50 = 1*V2V2 = 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
molmass of M2CO3 = 1 gmoles = 0.010 molmolar 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 4.

A quantity of 1 g of metal carbonate was dissolved in 25 ml normal HCl. The resulting liquid requires 50 ml of N/10 caustic soda solution to neutralize it completely. The equivalent weight of metal carbonate is:


To find the equivalent weight of metal carbonate, we need to use the formula:

Equivalent weight = Molecular weight / Acid or Base equivalent weight


Normality of HCl = [Number of H+ ions]/[Volume of solution in liters]

Since the solution is normal, the normality of HCl is 1 N.

Therefore, the number of H+ ions in 1 L of HCl = 1 N x 1 mole/L = 1 mole

The number of H+ ions in 25 ml of HCl = (1 mole/L) x (25 ml / 1000 ml) = 0.025 mole

HCl is a monoprotic acid, which means that it donates one H+ ion per molecule.

Therefore, the acid equivalent weight of HCl = Molecular weight / Number of H+ ions per molecule

= 36.5 g/mol / 1 = 36.5 g/mol


N/10 caustic soda solution means that the solution contains 1/10th of a mole of NaOH in 1 L of solution.

Therefore, the number of NaOH molecules in 50 ml of N/10 NaOH solution = (1/10) x (1/1000) x (50 ml) x (6.02 x 10^23 molecules/mol)
= 3.01 x 10^20 molecules

NaOH is a monobasic base, which means that it accepts one H+ ion per molecule.

Therefore, the base equivalent weight of NaOH = Molecular weight / Number of OH- ions per molecule

= 40 g/mol / 1 = 40 g/mol


The metal carbonate reacts with HCl as follows:

MCO3 + 2HCl → MCl2 + CO2 + H2O

From the above equation, we can see that 1 mole of metal carbonate reacts with 2 moles of HCl.

Therefore, the number of moles of HCl that reacted with 1 g of metal carbonate = 0.025 mole

The number of moles of metal carbonate in 1 g of metal carbonate = 1 g / Molecular weight

We do not know the molecular weight of the metal carbonate, so we cannot directly calculate the equivalent weight.

However, we know that the equivalent weight of metal carbonate is the same as its molecular weight, since it reacts with 2 moles of HCl to form 1 mole of the metal chloride.

Therefore, the equivalent weight of metal carbonate = Molecular weight of metal carbonate = 50 g/mol

Answer: d)50