60 g of glucose is dissolved in 250 g of water. Calculate the freezing point of this solution. (molar mass of glucose = $180~g~mol^{-1}$, $K_f$ for water = $1.86~K~kg~mol^{-1}$)
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Always convert the mass of the solvent to kilograms before calculating molality.
Step 1: Formula for freezing point depression. $\Delta T_f = K_f \times m$, where $m$ is the molality of the solution (moles of solute per kg of solvent). Step 2: Calculate moles of glucose. Molar mass of glucose $= 180~g~mol^{-1}$. Moles $= \frac{60}{180} = \frac{1}{3}~mol$. Step 3: Calculate molality. Mass of water $= 250~g = 0.250~kg$. Molality $m = \frac{1/3}{0.250} = \frac{4}{3} \approx 1.333~mol~kg^{-1}$. Step 4: Calculate freezing point. $\Delta T_f = 1.86 \times \frac{4}{3} = 2.48~K$. Freezing point $= 273.15 - 2.48 = 270.67~K$ (equivalent to $-2.48^\circ C$).