Value of kf of water is 1.86 k g mole ^-1 if your vehicles radiator can accommodate 1kg of water then how many grams of ethylene glycol can be dissolved in it so that freezing point of solution becomes -2.8 degree Celsius

Open in App

Solution

Normal freezing point of water = 0 degree celsius
Required freezing point = -2.8 degree celsius (given)
Depression in freezing point = 2.8 degrees
Volume of water = 1 kg of water
Kf of water = 1.86 kg mole ^-1

We have, depression in freezing point
ΔT = iK_fm

where m is molality and
i = 1 (ethylene glycol is an organic compound so it does not dissociate in solution and has an i factor of 1)
=> 2.8 = 11.86m
=> m= (2.8)/(1.86)
= 1.5053 g/L

Molality = no of moles of solute / Mass of solvent (in kg)
=> m = 1.5053 = no of moles of ethylene glycol / 1 kg
=> no of moles of ethylene glycol = 1.5053 moles

Mass of ethylene glycol that can be dissolved = no of moles * Molar mass
= 1.5053 * 62
= 93.33 grams

Suggest Corrections

Similar questions

K_{f}

1.86 K k g m o l^{- 1}

(C_{2} H_{6} O_{2})

- {2.8}^{o} C

(C_{2} H_{6} O_{2})

- {2.79}^{\circ} C

K_{f}

1.86 K k g m o l^{- 1}

Calculate the freezing point of a solution when 31 gram of Ethylene glycol is dissolved in 500 gram of water. (

Kf for water=1.86 k kg /mol

{103.57}^{o} C

(K_{f} = {1.86}^{o} C / m, K_{b} = {0.52}^{o} C / m)

- {0.3}^{0} C

K_{f}

K g m o l^{- 1}

Join BYJU'S Learning Program