Similar to For Practice 14.8) Determine the freezing point of an aqueous solution that contains 0.867 m glycerin (CHgOz).
Ki(water) = 1.86°C/m and Kg(water) - 0.512°C/m. Freezing point of water = 0.0 °C.

Similar to For Practice 14.3) Find the mass (in grams) of glucose (CH1206) in 505 mL of 10.5% glucose solution by mass. Assume the density of the solution is 1.04g/mL

The freezing point of the solution containing 0.867 m glycerin is -1.61442 °C. Option C is correct

The mass of glucose in 505 mL of 10.5% glucose solution is 53.01 g or 5.30 x 10^2 g.

Option C is correct

To find the freezing point depression of the solution containing 0.867 m glycerin:

ΔTf = Kf * molality

ΔTf = (1.86°C/m) * 0.867 m

ΔTf = 1.61442 °C

The freezing point depression is 1.61442 °C.

The freezing point of the solution is:

Freezing point = 0.0 °C - ΔTf

Freezing point = 0.0 °C - 1.61442 °C

Freezing point = -1.61442 °C

To find the mass of glucose in 505 mL of 10.5% glucose solution:

Mass of glucose = Volume of solution * Density of solution * % mass

Mass of glucose = 505 mL * 1.04 g/mL * 10.5%

Mass of glucose = 53.01 g

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You are in a laboratory creating a new chocolate bar. You want to create the sweetest chocolate bar by maximizing the sugar concentration. You are doing this by adding the sugar to a chocolate mixture. Which would allow you to dissolve more sugar?

The answer: Add the sugar after heating the mixture.