The stength of `10^(-2)Mna_2CO_3` solution interms of molality will be (density of the solution =`1.10gml^(-1))`
`(M.wtNa_2CO_3=106)`

Options

A. `9xx10^(-3)`
B. `1.15xx10^(-2)`
C. `5.1xx10^(-3)`
D. `11.2xx10^(-3)`

Answer

To find the strength of a \(10^{-2} M\) Na\(_2\)CO\(_3\) solution in terms of molality, we can use the following steps:

### Step 1: Understand the given data
- Molarity (M) of Na\(_2\)CO\(_3\) = \(10^{-2} M\)
- Density of the solution (D) = \(1.10 \, g/ml\)
- Molar mass of Na\(_2\)CO\(_3\) = \(106 \, g/mol\)

### Step 2: Use the formula to convert molarity to molality
The relationship between molality (m) and molarity (M) is given by the formula:

\[ m = \frac{M \times 1000}{D \times 1000 - Molar \, mass} \]

### Step 3: Substitute the values into the formula
Substituting the known values into the formula:

\[ m = \frac{10^{-2} \, \text{mol/L} \times 1000}{1.10 \, g/ml \times 1000 - 106 \, g/mol} \]

### Step 4: Calculate the denominator
Calculate the denominator:

\[ 1.10 \, g/ml \times 1000 = 1100 \, g/L \]

Now subtract the molar mass:

\[ 1100 \, g/L - 106 \, g/mol = 994 \, g/L \]

### Step 5: Substitute back into the equation
Now substitute back into the equation for molality:

\[ m = \frac{10^{-2} \times 1000}{994} \]

### Step 6: Perform the calculation
Calculating the numerator:

\[ 10^{-2} \times 1000 = 10 \]

Now divide by 994:

\[ m = \frac{10}{994} \approx 0.0101 \, mol/kg \]

### Step 7: Final answer
Thus, the molality of the solution is approximately:

\[ m \approx 9.9 \times 10^{-3} \, mol/kg \text{ or } 9 \times 10^{-3} \, molal \]