Question

A certain substance X has a normal freezing point of -10.1 degree C and a molal freezing point depression constant K_f = 5.32 degree C middot kg middot mol^-1. Calculate the freezing point of a solution made of 29.82 g of urea ((NH_2)_CO) dissolved in 500. g of X. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:
`-15.4^00C`

Step-by-step explanation:

`T^0_f-T_f=i* k_f* (w_2* 1000)/(M_2* w_1)`

where,

`T_f` = boiling point of solution = ?

`T^o_f` = boiling point of solvent (X) =
`-10.1^oC`

`k_f` = freezing point constant =
`5.32^oC/kgmol`

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte like urea)

= mass of solute (urea) = 29.82 g

= mass of solvent (X) = 500.0 g

`M_2` = molar mass of solute (urea) = 60 g/mol

Now put all the given values in the above formula, we get:

`(-10.1-(T_f))^oC=1* (5.32^oC/m)* ((29.82g)* 1000)/(60* (500.0g))`

`T_f=-15.4^0C`

Therefore, the freezing point of solution is
`-15.4^0C`