What is the g-molecular mass of nonionizing solid if 10 g of this solid dissolved in 100 g of water, forms a solution which froze at `-1.22^(@)` C ? `(K_(f)=1.86 K kg mol^(-1))`

Options

A. `265 g//mol`
B. `152 g//mol`
C. `130 g//mol`
D. `65 g//mol`

Explanation

To find the g-molecular mass of the non-ionizing solid, we can follow these steps:

### Step 1: Determine the depression in freezing point (ΔTf)
The freezing point of pure water is 0°C, and the freezing point of the solution is given as -1.22°C.

\[ \Delta T_f = T_f^{\text{pure}} - T_f^{\text{solution}} = 0°C - (-1.22°C) = 1.22°C \]

### Step 2: Calculate the molality (m) of the solution
Molality is defined as the number of moles of solute per kilogram of solvent.

1. **Calculate the number of moles of solute (solid):**
\[ \text{Number of moles} = \frac{\text{mass of solute (g)}}{\text{molar mass (M)}} \]
Given mass of solute = 10 g, so:
\[ \text{Number of moles} = \frac{10}{M} \]

2. **Convert the mass of water to kg:**
\[ \text{Mass of water} = 100 \text{ g} = 0.1 \text{ kg} \]

3. **Calculate molality (m):**
\[ m = \frac{\text{Number of moles}}{\text{mass of solvent (kg)}} = \frac{\frac{10}{M}}{0.1} = \frac{100}{M} \]

### Step 3: Use the freezing point depression formula
The relationship between the depression in freezing point, the cryoscopic constant (Kf), and molality is given by:
\[ \Delta T_f = K_f \cdot m \]
Substituting the known values:
\[ 1.22 = 1.86 \cdot \frac{100}{M} \]

### Step 4: Solve for M (molar mass)
Rearranging the equation to solve for M:
\[ M = \frac{1.86 \cdot 100}{1.22} \]

Calculating this gives:
\[ M \approx \frac{186}{1.22} \approx 152.46 \text{ g/mol} \]

### Conclusion
The g-molecular mass of the non-ionizing solid is approximately **152.46 g/mol**.

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