Question

Calculate the freezing point of the solution when 31 gram of ethylene glycol is dissolved in 500 gram of water

Answer 1

Hi there,

♤ Given :

● Ethylene glycol ( C2 H6 O2 )

Solute

● Its Molar mass = 12×2 + 6 + 16×2

= 62

● Its Sample mass = 31 gm

● Mass of water the solvent = 500gm = 0.5 Kg

● Kf = freezing constant of solvent

= 1.86c/m

♤ The solve

◇ Number of mols of solute =

sample mass ÷ molar mass

☆ = 31 ÷ 62 = 0.5 mol

◇ Molality of solution ( m ) =

Mols of solute ÷ mass of solvent in Kg

☆ 0.5 ÷ 0.5 = 1 m

☆ Delta T = Kf × m

T = 1.86 °c

Delta T = T solvent - T solution

◇ T solution = -1 86 °c

♡ Hope the answer is helpful

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