18 g of glucose $(C_{6}H_{12}O_{6})$ is dissolved in 1 kg of water, at what temperature will water boil at 1.013 bar? ($K_{b}$ for water = 0.52 K kg $mol^{-1}$)

Question

Accepted Answer

To determine the boiling point of the solution when 18 g of glucose is dissolved in 1 kg of water, we need to use the concept of boiling point elevation. The boiling point elevation can be calculated using the formula:

$$
\Delta T_b = i \cdot K_b \cdot m
$$

where:
- $\Delta T_b$ is the boiling point elevation.
- $i$ is the van't Hoff factor, which is 1 for glucose because it does not dissociate in solution.
- $K_b$ is the ebullioscopic constant for water, given as 0.52 K kg mol$^{-1}$.
- $m$ is the molality of the solution.

First, we calculate the molality ($m$) of the glucose solution:

1. Calculate the number of moles of glucose:
   - Molar mass of glucose ($C_6H_{12}O_6$) = 6(12) + 12(1) + 6(16) = 180 g/mol
   - Moles of glucose = $\frac{18 \text{ g}}{180 \text{ g/mol}} = 0.1 \text{ mol}$

2. Calculate the molality ($m$):
   - Molality ($m$) = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}}$
   - Molality ($m$) = $\frac{0.1 \text{ mol}}{1 \text{ kg}} = 0.1 \text{ mol/kg}$

Now, calculate the boiling point elevation ($\Delta T_b$):

$$
\Delta T_b = 1 \cdot 0.52 \text{ K kg mol}^{-1} \cdot 0.1 \text{ mol/kg} = 0.052 \text{ K}
$$

The normal boiling point of water is 373.15 K. Therefore, the new boiling point of the solution is:

$$
\text{New boiling point} = 373.15 \text{ K} + 0.052 \text{ K} = 373.202 \text{ K}
$$

Thus, the correct answer is Option 4: 373.202 K.

**Explanation for why other options are incorrect:**

1. **Option 1: 0.052 K** - This value represents the boiling point elevation ($\Delta T_b$), not the actual boiling point of the solution. It is the increase in temperature, not the final boiling point.

2. **Option 2: 0.52 K** - This value is incorrect because it does not correspond to either the boiling point elevation or the final boiling point. It seems to be a misinterpretation of the calculation involving $K_b$.

3. **Option 3: 373.15 K** - This is the normal boiling point of pure water at 1 atm (or 1.013 bar), not the boiling point of the solution. It does not account for the elevation caused by the dissolved glucose.

Therefore, based on the calculations and reasoning, Option 4 is indeed the correct answer.