Question

Kf for water is 1.86 K kg mol⁻¹. If your automobile radiator
holds 1.0 kg of water, how many grams of ethylene glycol
(C₂H₆O₂) must you add to get the freezing point of the
solution lowered to –2.8ºC ?
(a) 72 g (b) 93 g (c) 39 g (d) 27 g

Answer 1

Kf for water is 1.86 K kg mol⁻¹. If your automobile radiator

holds 1.0 kg of water, how many grams of ethylene glycol

(C₂H₆O₂) must you add to get the freezing point of the

solution lowered to –2.8ºC ?

- The correct answer is (b) 93 g

- molecular weight of erhylene glycol = 62 gm

- ∆Tf = 0-(-2.8) = 2.8 °C

- mass of solvent = 1 kg = 1000 gm

- ∆Tf = Kf × m

2.8 = 1.86 × (w × 1000)/ (62 × 1000)

w = (2.8 × 62 )/1.86

w = 93.33 gm (approx)

- so the correct answer is (b)

Answer 2

93 g of ethylene glycol must be added to get the freezing point of the

solution lowered to -2.8ºC

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f$ = Depression in freezing point

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

$K_f$ = freezing point constant = $1.86Kkg/mol$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (water)= 1.0 kg  

Molar mass of solute (ethylene glycol) =  62 g/mol

Mass of solute (ethylene glycol)  = x g

$0-(-2.8)^0C=1\times 1.86\times \frac{xg}{62g/mol\times 1.0kg}$

$x=93g$

Thus 93 g of ethylene glycol must be added to get the freezing point of the

solution lowered to -2.8ºC

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