After adding a solute freezing point of solution decreases to `-0.186 ^@C `. Calculate `DeltaT_b ` if `K_f = 1.86 ^@C`/ molal and `K_b = 0.52^@C`/ molal.

Options

A. 0.521
B. 0.0521
C. 1.86
D. 0.0186

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the depression in freezing point (ΔTf)
The depression in freezing point (ΔTf) is calculated using the formula:
\[ \Delta T_f = T_f^{\text{solvent}} - T_f^{\text{solution}} \]
Where:
- \(T_f^{\text{solvent}} = 0^\circ C\) (freezing point of pure water)
- \(T_f^{\text{solution}} = -0.186^\circ C\)

Substituting the values:
\[ \Delta T_f = 0 - (-0.186) = 0.186^\circ C \]

### Step 2: Calculate the molality (m) of the solution
Using the depression in freezing point, we can find the molality using the formula:
\[ \Delta T_f = K_f \cdot m \]
Where:
- \(K_f = 1.86^\circ C/\text{molal}\)

Rearranging the formula to find molality (m):
\[ m = \frac{\Delta T_f}{K_f} \]
Substituting the values:
\[ m = \frac{0.186}{1.86} \approx 0.1 \text{ molal} \]

### Step 3: Calculate the elevation in boiling point (ΔTb)
Now, we can calculate the elevation in boiling point using the formula:
\[ \Delta T_b = K_b \cdot m \]
Where:
- \(K_b = 0.52^\circ C/\text{molal}\)

Substituting the values:
\[ \Delta T_b = 0.52 \cdot 0.1 = 0.052^\circ C \]

### Final Answer
Thus, the elevation in boiling point (ΔTb) is:
\[ \Delta T_b = 0.052^\circ C \]

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