`4.2 g` of a metallic carbonate `MCO_(3)` was heated in a hard glass tube and `CO_(2)` evolved was found to have `1120 mL` of volume at `STP`. The `Ew` of the metal is

Options

A. 12
B. 24
C. 18
D. 15

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the number of moles of CO₂ evolved
Given that the volume of CO₂ evolved is 1120 mL at STP (Standard Temperature and Pressure), we know that 1 mole of gas occupies 22400 mL at STP.

\[ \text{Number of moles of } CO₂ = \frac{\text{Volume of } CO₂}{\text{Molar volume at STP}} = \frac{1120 \, \text{mL}}{22400 \, \text{mL/mol}} = 0.05 \, \text{moles} \]

### Step 2: Relate moles of CO₂ to moles of MCO₃
From the thermal decomposition of the metallic carbonate (MCO₃), we know that 1 mole of MCO₃ produces 1 mole of CO₂. Therefore, the moles of MCO₃ that decomposed is also 0.05 moles.

### Step 3: Calculate the molar mass of MCO₃
Using the given mass of MCO₃ (4.2 g) and the number of moles (0.05 moles), we can calculate the molar mass.

\[ \text{Molar mass of } MCO₃ = \frac{\text{mass}}{\text{moles}} = \frac{4.2 \, \text{g}}{0.05 \, \text{moles}} = 84 \, \text{g/mol} \]

### Step 4: Set up the equation for the molar mass of MCO₃
The molar mass of MCO₃ can be expressed in terms of the atomic mass of the metal (M) as follows:

\[ \text{Molar mass of } MCO₃ = \text{Atomic mass of } M + \text{Atomic mass of C} + 3 \times \text{Atomic mass of O} \]
\[ 84 = X + 12 + 3 \times 16 \]
\[ 84 = X + 12 + 48 \]
\[ 84 = X + 60 \]
\[ X = 84 - 60 = 24 \, \text{g/mol} \]

### Step 5: Calculate the equivalent weight of the metal
The equivalent weight (Ew) of the metal can be calculated using its atomic mass and its valency. Assuming the valency of the metal is 2 (common for metals in carbonates):

\[ \text{Equivalent weight} = \frac{\text{Atomic mass}}{\text{Valency}} = \frac{24 \, \text{g/mol}}{2} = 12 \, \text{g/equiv} \]

### Final Answer
The equivalent weight of the metal is **12 g/equiv**.

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