When `45gm` solution is dissolved in `600gm`water freezing point lower by `2.2K`,calcuate molar mass of solute `(K_(f)=1.86 "kg mol"^(-1))`

Options

A. `63.4`
B. `80gm`
C. `90gm`
D. `21gm`

Explanation

To calculate the molar mass of the solute, we can use the formula for freezing point depression:

\[ \Delta T_f = K_f \cdot m \]

where:
- \(\Delta T_f\) is the depression in freezing point,
- \(K_f\) is the cryoscopic constant of the solvent (water in this case),
- \(m\) is the molality of the solution.

### Step 1: Calculate the molality of the solution

Molality (m) is defined as the number of moles of solute per kilogram of solvent. We can express it as:

\[ m = \frac{n}{M_{solvent}} \]

where:
- \(n\) is the number of moles of solute,
- \(M_{solvent}\) is the mass of the solvent in kg.

Given:
- Mass of water (solvent) = 600 g = 0.6 kg,
- Mass of solute = 45 g.

To find the number of moles of solute (\(n\)), we need to use the molar mass (\(M\)) of the solute, which we are trying to find:

\[ n = \frac{\text{mass of solute}}{M} = \frac{45 \, \text{g}}{M} \]

Thus, the molality can be expressed as:

\[ m = \frac{n}{M_{solvent}} = \frac{45/M}{0.6} \]

### Step 2: Substitute molality into the freezing point depression formula

Now we can substitute this expression for molality into the freezing point depression formula:

\[ \Delta T_f = K_f \cdot m \]

Substituting the values we have:

\[ 2.2 = 1.86 \cdot \left(\frac{45/M}{0.6}\right) \]

### Step 3: Rearranging the equation to solve for M

Now we can rearrange the equation to solve for \(M\):

\[ 2.2 = 1.86 \cdot \frac{45}{0.6M} \]

Multiplying both sides by \(0.6M\):

\[ 2.2 \cdot 0.6M = 1.86 \cdot 45 \]

Calculating \(1.86 \cdot 45\):

\[ 2.2 \cdot 0.6M = 83.7 \]

Now divide both sides by \(2.2 \cdot 0.6\):

\[ M = \frac{83.7}{2.2 \cdot 0.6} \]

Calculating \(2.2 \cdot 0.6\):

\[ M = \frac{83.7}{1.32} \]

Calculating the final value:

\[ M \approx 63.4 \, \text{g/mol} \]

### Final Answer

The molar mass of the solute is approximately **63.4 g/mol**.

---