0.03 g of ethylene glycol (C2H6O7) must be added to 1.00 kg of water to produce a solution that freezes at -5.00C.
[tex]\Delta T_{f} = K_{f} m[/tex]
where,
ΔTf = Change in freezing point
Kf = molal freezing point depression constant of the solvent
m = moles of solute per kilogram of solvent.
Now,
[tex]\Delta T_{f} = K_{f} m[/tex]
[tex]\Delta T_{f} = K_{f} \times \frac{W_2}{M_2} \times \frac{1000}{W_1}[/tex] [-5.00°C = 273 + (-5.00) = 268 K] [tex]268 = 1.86 \times \frac{W_2}{62} \times \frac{1000}{0.001}[/tex] [1.00 kg = 0.001 g]
[tex]W_{2} = \frac{268 \times 62 \times 0.001}{1000 \times 1.86}[/tex]
= 0.03 g
Thus from the above conclusion we can say that 0.03 g of ethylene glycol (C2H6O7) must be added to 1.00 kg of water to produce a solution that freezes at -5.00C.
Learn more about the Freezing Point Depression here: https://brainly.com/question/24314907
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