The given problem involves the neutralization of a metal carbonate by hydrochloric acid. Let's break down the problem step by step to understand the concept and arrive at the correct answer.

Given:
- Amount of metal carbonate = 2 gm
- Volume of hydrochloric acid (HCl) = 100 ml
- Concentration of HCl = 0.1 N

To find: Equivalent weight of metal carbonate

Step 1: Understand the concept of equivalent weight
The equivalent weight of a substance is the amount of the substance that reacts with or is equivalent to 1 gram equivalent of hydrogen ions (H+). It is determined by the molar mass and the number of H+ ions involved in the reaction.

Step 2: Determine the number of moles of HCl used
To find the number of moles of HCl used, we can use the formula:

Moles = Concentration (N) × Volume (L)

Given that the concentration of HCl is 0.1 N and the volume used is 100 ml (which is equal to 0.1 L), we can calculate the number of moles of HCl:

Moles of HCl = 0.1 N × 0.1 L = 0.01 moles

Step 3: Determine the number of moles of metal carbonate
The reaction between metal carbonate and HCl can be represented as follows:

Metal Carbonate + 2HCl → Metal Chloride + H2O + CO2

From the balanced equation, we can see that 2 moles of HCl react with 1 mole of metal carbonate. Therefore, the number of moles of metal carbonate can be calculated as:

Moles of metal carbonate = 0.01 moles ÷ 2 = 0.005 moles

Step 4: Calculate the molar mass of metal carbonate
The molar mass of metal carbonate is the sum of the atomic masses of its constituent elements. Let's assume the metal carbonate is MCO3, where M represents the metal.

To find the molar mass, we need to know the atomic masses of M, C, and O. Let's assume the atomic masses of C and O are 12 g/mol and 16 g/mol, respectively.

Molar mass of MCO3 = Molar mass of M + Molar mass of C + (3 × Molar mass of O)
= M + 12 + (3 × 16)
= M + 60

Step 5: Calculate the equivalent weight of metal carbonate
The equivalent weight of metal carbonate is the molar mass divided by the number of moles involved in the reaction. In this case, the number of moles of metal carbonate involved in the reaction is 0.005 moles.

Equivalent weight of metal carbonate = Molar mass of MCO3 ÷ Moles of metal carbonate
= (M + 60) ÷ 0.005
= 200M + 1200

Since the equivalent weight is given as 200, we can equate it to the expression derived above:

200 = 200M + 1200
200M = -1000
M = -5

However, the value of M cannot be negative as it represents the atomic mass of the metal. Therefore, the correct answer is not provided among the given options.

In conclusion