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

The Correct Option is D

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 K_b:
\(\Delta T_b = K_b \cdot m\)
\(0.1 = K_b \times 0.1\)
Dividing both sides by 0.1:
\(1 = K_b\)
Therefore, the molal elevation constant (K_b) of the liquid is (D) 1 K kg/mol.

Answer 2

The formula for elevation in boiling point is:
\( \Delta T_b = K_b \cdot m \)
where:
\( \Delta T_b \) = 0.1°C (given)
Mass of glucose = 1.8 g
Molar mass of glucose (C₆H₁₂O₆) = 180 g/mol
Mass of solvent = 100 g = 0.1 kg

Molality (m) = \( \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1.8/180}{0.1} = 0.1 \, \text{mol/kg} \)

Now use the formula:
\( \Delta T_b = K_b \cdot m \Rightarrow 0.1 = K_b \cdot 0.1 \Rightarrow K_b = \frac{0.1}{0.1} = 1 \, \text{Kkg/mol} \)

Correct Answer: 1 K kg/mol

Answer 3

Given: 

• Mass of glucose = 1.8 g
• Molar mass of glucose (C₆H₁₂O₆) = 180 g/mol
• Mass of solvent = 100 g = 0.1 kg
• Elevation in boiling point, ΔT_b = 0.1 °C

Using the formula:
$$ \Delta T_b = K_b \cdot m $$ where \( m \) is the molality.

Molality \( m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1.8 / 180}{0.1} = \frac{0.01}{0.1} = 0.1 \, \text{mol/kg} \)

Now, using the boiling point elevation formula:
$$ 0.1 = K_b \cdot 0.1 $$

Solving for \( K_b \):
$$ K_b = \frac{0.1}{0.1} = 1 \, \text{K kg/mol} $$

Correct answer: 1 K kg/mol