Question

2.58 0.72 g of camphor in 32 g of acetone produces an
elevation of 0.25°C in the boiling point of acetone.
Calculate the molecular mass of camphor. (Ky for
acetone = 1.72 K kg mol)
Ans. 154.8 g mor​

Answer 1

mass of camphor = 0.72g

Let us consider that molecular mass of camphor is M g/mol.

mass of acetone (solvent ) = 32g

so, molality = mole of solute/mass of solvent in kg

= 0.72 × 1000/M × 32

= 720/32M

now using formula, ∆Tb =Kb × m

where , m = 720/32M , Kb = 1.72K kg/mol and ∆Tb = 0.25°C

so, 0.25 = 1.72 × 720/32M

or, M = 720 × 1.72/(32 × 0.25)

= 154.8 g/mol

Answer 2

Answer:

154.8g mol-1

Explanation:

To calculate the molecular mass of camphor let's use the given data:

Mass of the solvent (acetone) WA= 32g

Mass of the solute (camphor) WB= 0.72g

The Ky for acetone = 1.72 K Kg mol-1

Using the formula:

M = 1000 X Ky X WB/ TB X W

M = 1000 X 1.72 X 0.70/0.25 X 32

=154.8 g mol-1

So, the molecular mass of camphor will be 154.8g mol-1