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K for water is 1.86K 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 –2.8o C?

Option: 1

72g


Option: 2

93g


Option: 3

39g


Option: 4

27g


Answers (1)

best_answer

K_{f}=1.86\, K\, Kg\; mol^{-1}

\Delta T_{f}=0-(-2.8)=2.8^{\circ}C

Mass of solvent = 1.0 kg

Mass of solute = ?

Molecular mass of solute = 62

\Delta T_{f}=K_{f}\times m

m=\frac{w/62}{1000}\times 1000=\frac{w}{62}

\Delta T_{f}=K_{f}\times m

2.8=1.86\times \frac{w}{62}\Rightarrow w=\frac{62\times 2.8}{1.86}=93g

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avinash.dongre

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