40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is _________ K. (Nearest integer)
[Given : K$$_f$$ = 1.86 K kg mol$$^{-1}$$; Density of water = 1.00 g cm$$^{-3}$$; Freezing point of water = 273.15 K]

We start with the data given in the question. The mass of glucose added is 40 g and its molar mass is 180 g mol−1. The volume of water is 200 mL and the density of water is 1.00 g cm−3, so 200 mL of water has a mass of 200 g. The cryoscopic constant (freezing-point depression constant) of water is given as $$K_f = 1.86\;{\rm K\,kg\,mol^{-1}}$$ and the normal freezing point of pure water is 273.15 K.

First we calculate the number of moles of glucose present. The definition of amount of substance (moles) is

$$n = \frac{\text{mass}}{\text{molar mass}}.$$

Substituting the numerical values, we have

$$n_{\text{glucose}} = \frac{40\;{\rm g}}{180\;{\rm g\,mol^{-1}}} = \frac{40}{180}\;{\rm mol} = \frac{2}{9}\;{\rm mol} \approx 0.2222\;{\rm mol}.$$

Now we determine the mass of the solvent (water) in kilograms because molality is defined with kilograms of solvent:

$$m_{\text{water}} = 200\;{\rm g} = 0.200\;{\rm kg}.$$

The molality $$m$$ of the solution is given by the formula

$$m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}.$$

Substituting the values, we get

$$m = \frac{0.2222\;{\rm mol}}{0.200\;{\rm kg}} = 1.111\;{\rm mol\,kg^{-1}}.$$

Next, the freezing-point depression $$\Delta T_f$$ is calculated using the relation

$$\Delta T_f = i\,K_f\,m,$$

where $$i$$ is the van ’t Hoff factor. For glucose, which is a non-electrolyte and does not dissociate, $$i = 1.$$ Therefore

$$\Delta T_f = (1)\,(1.86\;{\rm K\,kg\,mol^{-1}})\,(1.111\;{\rm mol\,kg^{-1}}).$$

Carrying out the multiplication,

$$\Delta T_f = 1.86 \times 1.111 \approx 2.066\;{\rm K}.$$

The new freezing point $$T_{\text{solution}}$$ is obtained by subtracting the depression from the normal freezing point of water:

$$T_{\text{solution}} = 273.15\;{\rm K} - 2.066\;{\rm K} \approx 271.084\;{\rm K}.$$

On rounding to the nearest integer, the freezing point becomes 271 K.

So, the answer is $$271\;{\rm K}.$$