Compute the molality of solute particles and the freezing temperature of each of

Question

Compute the molality of solute particles and the freezing temperature of each of the following solutions 1. 5.90 g of CoH1206 (a non-electrolyte) in 250 g of ethanol a. b. 14.66 g of MgF2 (a strong electrolyte) in 0.30 kg of water 2. A solution is prepared by mixing 15.0 g of NHACI (a strong electrolyte) in 500 g of water. Determine the molality and boiling point of this solution. 3. State the solute-solvent interaction involved in each of the following solutions: a. NaBr in water b. O2 in CCl4 c. CO in water d. CH3OH in water

Explanation / Answer

1)
a)

Molar mass of C6H12O6,
MM = 6*MM(C) + 12*MM(H) + 6*MM(O)
= 6*12.01 + 12*1.008 + 6*16.0
= 180.156 g/mol


mass(C6H12O6)= 5.90 g

use:
number of mol of C6H12O6,
n = mass of C6H12O6/molar mass of C6H12O6
=(5.9 g)/(180.156 g/mol)
= 3.275*10^-2 mol

m(solvent)= 250 g
= 0.25 kg

use:
Molality,
m = number of mol / mass of solvent in Kg
=(3.275*10^-2 mol)/(0.25 Kg)
= 0.131 molal
Answer: 0.131 molal

b)
Molar mass of MgF2,
MM = 1*MM(Mg) + 2*MM(F)
= 1*24.31 + 2*19.0
= 62.31 g/mol


mass(MgF2)= 14.66 g

use:
number of mol of MgF2,
n = mass of MgF2/molar mass of MgF2
=(14.66 g)/(62.31 g/mol)
= 0.2353 mol

m(solvent)= 0.30 Kg

use:
Molality,
m = number of mol / mass of solvent in Kg
=(0.2353 mol)/(0.3 Kg)
= 0.784 molal
Answer: 0.784 molal

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