Kf r for water is 1.86Kg 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 hte solution lowered to – 2.8 °C?

# ${K}_{\mathrm{f}}$ r for water is . 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 hte solution lowered to - 2.8 °C?

1. A

72 g

2. B

93 g

3. C

39 g

4. D

27 g

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### Solution:

Since $-\mathrm{\Delta }{T}_{\mathrm{f}}={K}_{\mathrm{f}}m$ we get

$m=\frac{-\mathrm{\Delta }{T}_{\mathrm{f}}}{{K}_{\mathrm{f}}}=\frac{2.8\mathrm{K}}{1.86\mathrm{Kkg}{\mathrm{mol}}^{-1}}=1.505\mathrm{mol}$

Since the molality, $m=n/{m}_{1}$, we get

$n=m{m}_{1}=\left(1.505\mathrm{mol}{\mathrm{kg}}^{-1}\right)\left(1\mathrm{kg}\right)=1.505\mathrm{mol}$

Finally, the mass of ethylene glycol, ${\mathrm{C}}_{2}{\mathrm{H}}_{6}{\mathrm{O}}_{2}$ (the molar mass = 62 g mor -1) required will be

$m=n{M}_{\mathrm{m}}=\left(1.505\mathrm{mol}\right)\left(62\mathrm{g}{\mathrm{mol}}^{-1}\right)=93.3\mathrm{g}$  +91

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