YOU dissolve 0.0100 g of ethylene glycol (HOCH2CH2OH) at 25 C in your sample of

Question

YOU dissolve 0.0100 g of ethylene glycol (HOCH2CH2OH) at 25 C in your sample of 10.00 g of diethyl ether at 1 atm.

A) assume there is no delta V mixing when you prepare the solution.

I) molality; m

ii) ppm of ethylene glycol

iii) Molarity, M assuming a density of solution at 0..7134 g/ mL.

IV) THEN YOU PIPPETE 3 mL from the solution and dilute to 50.00 mL IN A VOLUMETRIC FLASK. what is the initial Molarity ?( hint use M in IIIabove as concentrate M)

B) the new vapor pressure 17.9 celsisus when 0.01000 g of ethylene glycol is dissolved

c) the new boiling point at 1 atm , Kb= 2.02 (Celsius/ m) when 0.01000 g of ethylene glycol is dissolved.

D) the new freezing pt at 1 atm, Kf= 1.79 9celsius/m) when 0.01000 g of ethylene glycol is dissolved.

Explanation / Answer

Q1.

i) molality = mol of solute / kg solvent

mol of solute = mass/MW = 0.01/62.07 = 0.00016110 mol

kg solvent = 10 g = 10*10^-3 = 0.01 kg

molal = 0.00016110 /0.01 = 0.01611 molal

ii=

ppm = mg of EG / kg of EG = (0.01*10^-3) gm / (10^-3) kg = 0.001 ppm

iii)

M = mol / V

V = mass/D = (0.01+10)/(0.7134) = 14.0313 mL = 14.0313*10^-3 L

M = (0.00016110)/( 14.0313*10^-3)= 0.011481 mol per liter

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