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

1 K/m

The correct answer is option 1. 1 K/m.

To calculate the molal elevation constant of the liquid, we can use the formula:

\(\Delta T_b = K_b \cdot m\)

where \(\Delta T_b\) is the rise in boiling point, \(K_b\) is the molal elevation constant, and \(m\) is the molality of the solution.

Given that the rise in boiling point (\(\Delta T_b\)) is 0.1°C and the mass of glucose is 1.8 g, we need to find the molality (\(m\)) of the solution. The molality is defined as:

\(m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\)

First, we calculate the number of moles of glucose:

\(\text{moles of glucose} = \frac{\text{mass of glucose}}{{\text{molar mass of glucose}}} = \frac{1.8 \, \text{g}}{180 \, \text{g/mol}} = 0.01 \, \text{mol}\)

Next, we calculate the mass of the solvent in kg:

\(\text{mass of solvent} = \text{total mass} - \text{mass of solute} = 100 \, \text{g} - 1.8 \, \text{g} = 98.2 \, \text{g} = 0.0982 \, \text{kg}\)

Now, we can calculate the molality:

\(m = \frac{0.01 \, \text{mol}}{0.0982 \, \text{kg}} \approx 0.102 \, \text{mol/kg}\)

Finally, we can determine the molal elevation constant (\(K_b\)) by rearranging the equation:

\(K_b = \frac{m \cdot \Delta T_b}{0.1^\circ \text{C}} = \frac{0.102 \, \text{mol/kg}}{0.1^\circ \text{C}} \approx 1 \, \text{K/m} \)

Therefore, the molal elevation constant of the liquid is approximately 0.98 K/m, which corresponds to option (1).