QJEEM23Ch9962: JEE Main 2023 Chemistry - Class 12 JEEM23Ch9962 Solution: What weight of glucose must be dissolved in of water to lower…

The lowering of vapor pressure of a solvent by a nonvolatile solute is given by Raoult's law:

$\Delta P = x_{\text{solute}} \cdot P_0$

where $\Delta P$ is the change in vapor pressure, $x_{\text{solute}}$ is the mole fraction of the solute, and $P_0$ is the vapor pressure of the pure solvent.

Rearranging the formula for $x_{\text{solute}}$ , we have:

$x_{\text{solute}} = \frac{\Delta P}{P_0}$

Substituting the given values, we get :

$x_{\text{solute}} = \frac{0.20 \, \text{mm Hg}}{54.2 \, \text{mm Hg}} = 0.003689$

Since the mole fraction of the solute is also equal to the number of moles of solute divided by the total number of moles, we can express $x_{\text{solute}}$ as:

$x_{\text{solute}} = \frac{\text{moles}_{\text{solute}}}{\text{moles}_{\text{solute}} + \text{moles}_{\text{water}}}$

Assuming that the solution is dilute, the number of moles of water will be much larger than the number of moles of solute, so we can approximate the total number of moles as the number of moles of water.

Thus, we have :

$x_{\text{solute}} \approx \frac{\text{moles}_{\text{solute}}}{\text{moles}_{\text{water}}}$

Therefore, the number of moles of solute is :

$\text{moles}_{\text{solute}} = x_{\text{solute}} \cdot \text{moles}_{\text{water}}$

The number of moles of water is the mass of the water divided by the molar mass of water (18 g/mol) :

$\text{moles}_{\text{water}} = \frac{100 \, \text{g}}{18 \, \text{g/mol}} = 5.56 \, \text{mol}$

So, the number of moles of solute is :

$\text{moles}_{\text{solute}} = 0.003689 \cdot 5.56 \, \text{mol} = 0.0205 \, \text{mol}$

Finally, to find the mass of the glucose, we multiply the number of moles of glucose by the molar mass of glucose :

$\text{mass}_{\text{glucose}} = \text{moles}_{\text{solute}} \cdot \text{molar mass}_{\text{glucose}}$

$\text{mass}_{\text{glucose}} = 0.0205 \, \text{mol} \cdot 180 \, \text{g/mol} = 3.69 \, \text{g}$

So, the correct answer is 3.69 g.