The molal freezing point depression constant =kf6.19·°c·kgmol−1 for a certain substance x . when 10.g of urea nh22co are dissolved in 500.g of x , the solution freezes at −6.8°c . calculate the freezing point of pure x . be sure your answer has the correct number of significant digits.

-4.7°C

Explanation:

The freezing point depression is a colligative property that can be calculated using the following expression.

ΔTf = Kf . b

where

ΔTf: depression in the freezing point

Kf: molal freezing point depression constant

b: molality

The moles of urea (solute) are:

The mass of solvent (X) is 500 g (0.500 kg).

The molality of the urea is:

ΔTf = Kf . b = (6.19 °C.kg/mol) × 0.34 mol/kg = 2.1°C

The freezing point of pure X is:

-6.8°C + 2.1°C = -4.7°C

- lead + phosphoric acid = hydrogen + lead(ii) phosphate

1. they vary with a higher number because boiling is way hotter than melting