Question

An aqueous solution containing 16.5 g of an unknown molecular (nonelectrolyte) compound in 106.0 g of water was found to have a freezing point of -1.9 ∘C. Calculate the molar mass of the unknown compound.

Answer 1

Answer: The molar mass of the unknown compound is 152 g/mol.

Step-by-step explanation:

Depression in freezing point is given by:

`\Delta T_f=i* K_f* m`

`\Delta T_f=T_f-^0T_f=(0-(-1.9)^0C=1.9^0C` = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

`K_f` = freezing point constant =
`1.86^0C/m`

m= molality

`\Delta T_f=i* K_f* \frac{\text{mass of solute}}{\text{molar mass of solute}* \text{weight of solvent in kg}}`

Weight of solvent (water)= 106.0 g = 0.106 kg

Molar mass of unknown non electrolyte = M g/mol

Mass of unknown non electrolyte added = 16.5 g

`1.9=1* 1.86* (16.5g)/(M g/mol* 0.106kg)`

`M=152g/mol`

Thus the molar mass of the unknown compound is 152 g/mol.

Answer 2

Answer : The molar mass of unknown compound is 152.38 g/mole

Explanation :

Depression in freezing point =
`1.9^oC`

Mass of unknown compound = 16.5 g

Mass of water = 106.0 g

Formula used :

`\Delta T_f=i* K_f* m\\\\\Delta T_f=i* K_f*\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}* \text{Mass of water in Kg}}`

where,

`\Delta T_f` = depression in freezing point

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

`K_f` = freezing point constant for water =
`1.86^oC/m`

m = molality

Now put all the given values in this formula, we get

`1.9^oC=1* (1.86^oC/m)* \frac{16.5g* 1000}{\text{Molar mass of unknown compound}* 106.0g}`

`\text{Molar mass of unknown compound}=152.38g/mole`

Therefore, the molar mass of unknown compound is 152.38 g/mole