Question

When 12.1 g of the sugar sucrose (a nonelectrolyte) are dissolved in exactly 800 g of water, the solution has a freezing point of –0.082°C. What is the molar mass of sucrose? [Kf of water is 1.86°C/m.]

Answer 1

Answer: 343g/mol

Step-by-step explanation:

Depression in freezing point is given by:

`\Delta T_f=K_f* m`

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

`\Delta T_f=T_f^0-T_f=(0-(-0.082)^0C=0.082^0C` = Depression in freezing point

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

m= molality

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

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

`0.082^0C=1* (1.86^0C/kgmole)* \frac{12.1g}{\text{Molar mass of sucrose}* 0.8kg}`

`\text{Molar mass of sucrose}=343g/mol`

Thus the molar mass of sucrose is 343 g/mol