Question

The rise in boiling point of a solution containing 1.8 g of glucose in 100 g of solvent is 0.1 degree C. The molal elevation constant of the liquid is

• A. 2 K kg/mol
• B. 0.1 K kg/mol
• C. 10 K kg/mol
• D. 1 K kg/mol

Answer 1

Answer: (D)

1 K kg/mol

Answer 2

To calculate the molal elevation constant (Kb) of the liquid, we can use the formula:
$\Delta T_b = K_b \cdot m$
where $\Delta T_b$ is the elevation in boiling point and m is the molality of the solution.
Given that the rise in boiling point ($\Delta T_b$) is 0.1 degree Celsius and the mass of the solvent is 100 g, we need to determine the molality of the solution.
The molality (m) can be calculated using the formula:
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$
To find the moles of solute, we need to convert the mass of glucose to moles using its molar mass.
The molar mass of glucose $\text{C}_6\text{H}_{12}\text{O}_6$ is 180 g/mol.

$\text{Moles of glucose} = \frac{\text{Mass of glucose}}{\text{Molar mass of glucose}}$

= $\frac{1.8 \, \text{g}}{180 \, \text{g/mol}}$
$= 0.01 mol$

Now we can calculate the molality (m):
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$

= $\frac{0.01 \, \text{mol}}{0.1 \, \text{kg}}$
= 0.1 mol/kg

Now we can substitute the values into the equation to find Kb:
$\Delta T_b = K_b \cdot m$
$0.1 = K_b \times 0.1$
Dividing both sides by 0.1:
$1 = K_b$
Therefore, the molal elevation constant (Kb) of the liquid is (D) 1 K kg/mol.