A solution is made by dissolving 1.00 $\times$ 10² g of essentially nonvolatile ethylene glycol (C₂H₆O₂)in 5.00 $\times$ 10² g of water.What is the resulting freezing point of the solution? (K_f = 1.86 $\circ$ C/m)

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Explanation:The freezing point depression is given by the equation:$$\Delta T_f = K_f 	imes m$$where:$$\Delta T_f$$ is the freezing point depression$$K_f$$ is the freezing point depression constant$$m$$ is the molality of the solutionThe molality of the solution is given by:$$m = \frac{moles\ of\ solute}{kg\ of\ solvent}$$The number of moles of ethylene glycol is:$$moles\ of\ ethylene\ glycol = \frac{100\ g}{62.07\ g/mol} = 1.61\ mol$$The mass of the solvent (water) is:$$mass\ of\ water = 500\ g = 0.5\ kg$$Therefore, the molality of the solution is:$$m = \frac{1.61\ mol}{0.5\ kg} = 3.22\ mol/kg$$The freezing point depression is then:$$\Delta T_f = 1.86\ ^\circ C/m 	imes 3.22\ mol/kg = -5.99\ ^\circ C$$The freezing point of the solution is therefore:$$-5.99\ ^\circ C$$

A Solution Is Made by Dissolving 1 ×\times× 102 G of Essentially Nonvolatile Ethylene Glycol (C2H6O2)in 5

A Solution Is Made by Dissolving 1 $\times$ 10² G of Essentially Nonvolatile Ethylene Glycol (C₂H₆O₂)in 5