Question

At −11.0 ∘C , a common temperature for household freezers, what is the maximum mass of sucrose (C12H22O11) you can add to 3.00 kg of pure water and still have the solution freeze? Assume that sucrose is a molecular solid and does not ionize when it dissolves in water.

Answer 1

6.064 kg

Step-by-step explanation:

The freezing point depression of water is 1.86° C/m

To still freeze at -11.0° C, the molarity of the solution has to be no more than
`\frac{11.0} {1.86}` = 5.91 m.

The molecular weight of sucrose is:

12 C = 12 *12 = 144

22 H = 22 * 1 = 22

11 O = 11 * 16 = 176

Total = 342 g/mol

Thus one mole of sucrose has a mass of 342 grams.

However, since there are 3.00 kg of water, you can add
`3.00 * 5.91 moles`(= 17.73 moles) to the water to get the -11.0° C freezing point depression.

Finally;

mass = number of mole * molar mass

= 17.73 moles * 342 grams / mole = 6063.66 grams

= 6.064 kg