Question

when 1.80g of a non volatile compound are dissolved in 25.0g of acetone, the solution boils at 56.86°C while pure acetone boils at 56.38°C under the same atmospheric pressure. calculate the molecular weight of the compound. The molal elevation constant for acetone is 1.72 K Kg mol-1​

Answer 1

Given info : when 1.80g of a non volatile compound are dissolved in 25.0g of acetone, the solution boils at 56.86°C while pure acetone boils at 56.38°C under the same atmospheric pressure. The molal elevation constant for acetone is 1.72 K kg/mol

To find : The molecular weight of the compound is...

solution : molality, m = no of moles of solute/mass of solvent in kg

= (1.8/M)/(25/1000)

= 72/M

where M is molecular weight of solute.

now using formula, ∆Tb = K_b × m

here K_b is molal elevation constant, m is molality and ∆T_b is elevation of boiling point.

∆T_b = 56.86°C - 56.38°C = 0.48°C

m = 72/M

K_b = 1.72 K kg/mol

∴ 0.48 = 1.72 × 72/M

⇒M = 72 × 1.72/0.48 = 72 × 4 = 288 g/mol

Therefore the molecular weight of the compound is 288 g/mol.

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Answer 2

Answer:

this is the required answer