$\mathrm{K}_{\mathrm{f}}$ for water is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol $\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)$ must you add to get the freezing point of the solution lowered to $-2.8^{\circ} \mathrm{C}$ ?

JEE Main · 2012 · 29 Apr 2012 · Offline · Solutions

K_{f}

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

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

must you add to get the freezing point of the solution lowered to

- 2. 8^{\circ} C

(A)72 g
(B)93 g✓ Correct
(C)39 g
(D)27 g

Answer: (B)

Solution

The freezing point is lowered by

Δ T_{f} = K_{f} m

. Here

Δ T_{f} = 2.8

K and

K_{f} = 1.86 K kg mo l^{- 1}

, so the molality required is

m = \frac{2.8}{1.86} \approx 1.505 mol/kg

. With

1.0

kg of water that is

1.505

mol of ethylene glycol, and at a molar mass of

62 g/mol

for

C_{2} H_{6} O_{2}

the mass needed is

1.505 \times 62 \approx 93

JEE Main · 2012 · 29 Apr 2012 · Offline · Single correct option

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