Question

When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°C less than the freezing point of pure X.Calculate the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point. The van't Hoff factor =i1.72 for potassium bromide in X.

Answer 1

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Step-by-step explanation:

Depression in freezing point is given by:

`\Delta T_f=i* K_f* m`

`\Delta T_f=T_f^0-T_f=8.30^0C` = Depression in freezing point

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

`K_f` = freezing point constant =

m= molality =
`\frac{\text{mass of solute}}{\text{molar mass of solute}* \text{weight of solvent in kg}}=(74.8g* 1000)/(1450g* 89.09g/mol)=0.579`

`8.30^0C=1* K_f* 0.579`

`K_f=14.3^0C/m`

Let Mass of solute (KBr) = x g

`8.3^0C=1.72* 14.3* (xg* 1000)/(119g/mol* 1450g)`

`x=58.2g`

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams