Question

A certain liquid has a normal freezing point of and a freezing point depression constant . A solution is prepared by dissolving some glycine () in of . This solution freezes at . Calculate the mass of that was dissolved.

Answer 1

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of
`-6.4^oC` and a molal freezing point depression constant
`K_f= 3.96^oC.kg/mol`. A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at
`-13.6^oC`. Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

Answer: 129.66 g of glycine will be dissolved.

Step-by-step explanation:

Depression in the freezing point is the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

`\text{Freezing point of pure solvent}-\text{freezing point of solution}=i* K_f* m`

OR

....(1)

where,

i = Van't Hoff factor = 1 (for non-electrolytes)

Freezing point of pure solvent =
`-6.4^oC`

Freezing point of solution =
`-13.6^oC`

`K_f` = freezing point depression constant =
`3.96^oC/m`

`M_(solute)` = Molar mass of solute (glycine) = 75.07 g/mol

`w_(solvent)` = Mass of solvent = 950 g

Plugging values in equation 1:

`-6.4-(-13.6)=1* 3.96* \frac{\text{Given mass of glycine}* 1000}{75.07* 950}\\\\\text{Given mass of glycine}=(7.2* 75.07* 950)/(1* 3.96* 1000)\\\\\text{Given mass of glycine}=129.66g`

Hence, 129.66 g of glycine will be dissolved.