COLLIGATIVE PROPERTES 1. What is the e molality of the solution prepared by diss

Question

COLLIGATIVE PROPERTES 1. What is the e molality of the solution prepared by dissolving 12.5 g of ethyl alcohol, C2HsOH, in 750. mL of water 1.25 g of sulfur in 250. mL of benzene (density of benzene = 0.875 g/mL) A solution is prepared by dissolving 88.5 g of sugar, CeH120s, in 300. mL of water. What is the vapor pressure of the solution? Vapor pressure of pure water is 28 torr at that temperature. 3. What is the boiling point of a solution prepared by dissolving 0.144 g of oil of clove, C10H12O2, in 10.0 g of benzene? K.-2.53 "C/m and the boiling point of pure benzene is 80.10 °C. 4. Calculate the molar mass of oil of wintergreen from the following data: 1.25 g of oil is dissolved in 100. g of benzene. The boiling point of the solution is 80.31 Boiling point of pure benzene is 80.10 C, K.-2.53C/m. How many grams of a non-ionic solid is dissolved in 275 mL of water if the molar mass of the solid is 80 g/mole, Ky -1.86 °C/m and the solution freezes at -0.32 °C ? 5. A 0.00200m aqueous solution of an ionic substance freezes at 0.00732 . How many ions does one mole of this substance give on being dissolved? Kf 1.86 C/m 6. List the following aqueous solutions in the order of increasing boiling point: a) 0.120m sugar b) 0.050m LiBr c) 0.045m NazSO 7.

Explanation / Answer

Q1.

a)

molality = mol of solute / kg of solvent

mol of alcohol = mass/MW = (12.5/46) = 0.2717

mass of water = 750 mL --> 750 g approx = 0.75 kg

molality =0.2717/0.75

molality = 0.3622

b)

mol of S = mass/MW = 7.25/32 = 0.2265

mass solvent = D*V = 0.875*250 = 218.75 g = 0.221875 kg

molality = mol/kg = 0.2265/0.221875 = 1.02 approx

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