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

Explanation

To calculate the molar mass of the solute, we can use the formula for the depression in freezing point, which is given by:

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

Where:
- \(\Delta T_f\) = depression in freezing point (in Kelvin)
- \(K_f\) = cryoscopic constant (in kg/mol)
- \(m\) = molality of the solution (in mol/kg)

### Step 1: Calculate the molality (m)

Molality (m) is defined as the number of moles of solute per kilogram of solvent. It can be calculated using the formula:

\[ m = \frac{W_b}{M_b} \cdot \frac{1000}{W_a} \]

Where:
- \(W_b\) = mass of solute (in grams)
- \(M_b\) = molar mass of solute (in grams/mol)
- \(W_a\) = mass of solvent (in grams)

### Step 2: Rearranging the equation

We can rearrange the equation for molality to find the molar mass of the solute:

\[ M_b = \frac{K_f \cdot W_b \cdot 1000}{\Delta T_f \cdot W_a} \]

### Step 3: Substitute the known values

Given:
- \(K_f = 1.86 \, \text{kg/mol}\)
- \(W_b = 45 \, \text{g}\)
- \(W_a = 600 \, \text{g}\)
- \(\Delta T_f = 2.2 \, \text{K}\)

Substituting these values into the rearranged equation:

\[ M_b = \frac{1.86 \cdot 45 \cdot 1000}{2.2 \cdot 600} \]

### Step 4: Calculate the molar mass

Now we will calculate the molar mass:

1. Calculate the numerator:
\[ 1.86 \cdot 45 \cdot 1000 = 83700 \]

2. Calculate the denominator:
\[ 2.2 \cdot 600 = 1320 \]

3. Now divide the numerator by the denominator:
\[ M_b = \frac{83700}{1320} \approx 63.41 \, \text{g/mol} \]

### Step 5: Conclusion

The molar mass of the solute is approximately \(63.4 \, \text{g/mol}\).

### Final Answer

The correct option for the molar mass of the solute is **Option A: 63.4 g/mol**.

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