A solution was prepared by dissolving 0.834 g of sulfur, S_t, in 100.0 g of acet
ID: 960810 • Letter: A
Question
A solution was prepared by dissolving 0.834 g of sulfur, S_t, in 100.0 g of acetic acid, CH_3COOH, calculate the freezing point and the boiling point of the solution..834 g Sulfur 100.0 g CH_3COOH freezing point: 14.48 degree C boiling point: 118.6 degree C Vanillin, C_8H_8O_3, occurs naturally in vanilla extract. A 39.1 mg sample of vanillin was dissolved in 168.5 mg of diphenyl ether (C_6H_5)_2O. What is the molality of the vanillin in the solution? m = mol solute/mass solvent = 39.1 mg right arrow 168.5 mg C_8H_8O_3 (C_6H_5)_2 O m = 168.5 mg/39.1 = 4.309 Calculate the mass percentage of Na_2SO_4 in a solution containing 10.6 g Na_2SO_4 in 48 g of H_2O water. Na_2SO_4 = (22.99)_2 + 32.066 + (16)_4 141.86 508.8/141.86 = 3.59 35.9%Explanation / Answer
1.
moles S8 = 0.834 g / 256.528 g/mol = 0.00325 moles
molality = (0.00325 moles) / (0.10 Kg) =0.0325 m
delta Tf = Kf x m
Kf of acetic acid = 3.9
delta Tf = (3.9) x (0.0325) = 0.126
freezing point = 16.6 - 0.126 =16.474
delta Tb = Kb x m
Kb of acetic acid = 3.07
delta Tf = (3.07) x (0.0325) = 0.01
boiling point = 118.1 + 0.01 = 118.11 °Cmolality = 0.00342 / 0.110 Kg =0.0310
delta Tf = 0.0310 x 3.90 = 0.121
freezing point = 16.5 - 0.121 =16.4 °C
delta Tb = 0.0310 x 3.07 = 0.0952
boiling point = 118.1 + 0.0952 = 118.2 °C
2.
molality (m) = moles solute / kg of solution
molar mass of Vanillin = 152.15 g/mol
So, 152.15 g of Vanillin = 1 mole
39.1 mg (=0.0391 g) of Vanillin = (0.0391 g / 152.15) mole = 0.000257 mole (solute)
168.5 mg of diphenyl ether = 0.0001685 kg of diphenyl ether (solvent)
So molality of of vanillin in the solution = (0.000257 mole) / (0.0001685 kg) 1.53 m
3.
% Mass = (mass solute / total mass solution) x 100
So, % mass Na2SO4 = 10.6 g / (10.6 g + 48 g) x 100
= 18.09 %
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