The concentration of a 100 ml solution containing X grams of `Na_(2)CO_(3)` (mol. wt. = 106) is Y M. The values of X and Y are respectively :

Options

A. `2.12, 0.05`
B. `1.06, 0.2`
C. `1.06, 0.1`
D. `2.12, 0.1`

Explanation

To solve the problem, we need to determine the values of X (grams of Na₂CO₃) and Y (molarity of the solution) based on the given information.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- We have a 100 ml solution containing X grams of Na₂CO₃.
- The molar mass of Na₂CO₃ is 106 g/mol.
- We need to find the values of X and Y, where Y is the molarity of the solution.

2. **Convert Volume from ml to L**:
- The volume of the solution is given as 100 ml.
- To convert ml to liters, we divide by 1000:
\[ \text{Volume in liters} = \frac{100 \text{ ml}}{1000} = 0.1 \text{ L} \]

3. **Calculate the Number of Moles of Na₂CO₃**:
- The number of moles of Na₂CO₃ can be calculated using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
- Substituting the values, we have:
\[ \text{Number of moles} = \frac{X}{106} \]

4. **Use the Molarity Formula**:
- The formula for molarity (Y) is given by:
\[ Y = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]
- Substituting the values we have:
\[ Y = \frac{\frac{X}{106}}{0.1} \]
- Simplifying this gives:
\[ Y = \frac{X}{106} \times 10 = \frac{10X}{106} = \frac{X}{10.6} \]

5. **Establish the Relationship Between X and Y**:
- From the equation \( Y = \frac{X}{10.6} \), we can express X in terms of Y:
\[ X = 10.6Y \]

6. **Determine the Values of X and Y**:
- We need to find specific values for X and Y. If we assume Y = 0.1 M (a common molarity for solutions), we can calculate:
\[ X = 10.6 \times 0.1 = 1.06 \text{ grams} \]
- Thus, we have:
- \( X = 1.06 \) grams
- \( Y = 0.1 \) M

### Final Answer:
The values of X and Y are:
- \( X = 1.06 \) grams
- \( Y = 0.1 \) M