In order to decrease the freezing point of 500.g of water to 1.0°C,how many grams of ethylene glycol (C₂H₆O₂)must be added? (K_f = 1.86 ) A)5.37 g B)1.08 g C)16.7 g D)66.7 g

Question

Quizplus · Accepted Answer

We can use the formula:

ΔT = Kf × molality

where ΔT is the change in freezing point (in °C), Kf is the freezing point depression constant (in °C/m), and molality is the concentration of the solute (in mol/kg).

Given that ΔT = 1.0°C, Kf = 1.86°C/m, and the mass of water is 500 g, we can calculate the molality of ethylene glycol as:

1.0 = 1.86 × (molality) 
molality = 1.0 / 1.86 = 0.537 mol/kg

The molality is the same as the number of moles of ethylene glycol per kilogram of water. To calculate the mass of ethylene glycol needed, we can use its molar mass (62.07 g/mol) as follows:

0.537 mol/kg × 62.07 g/mol × 500 g = 16.7 g

Therefore, the answer is choice C, 16.7 g of ethylene glycol.

In Order to Decrease the Freezing Point of 500