1. The common laboratory solvent benzene is often used to purify substances diss

Question

1. The common laboratory solvent benzene is often used to purify substances dissolved in it. The vapor pressure of benzene , C6H6, is 73.03 mm Hg at 25 °C. In a laboratory experiment, students synthesized a new compound and found that when 13.86 grams of the compound were dissolved in 182.1 grams of benzene, the vapor pressure of the solution was 71.47 mm Hg. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight of this compound? benzene = C6H6 = 78.12 g/mol. MW = ? g/mol

2.The boiling point of chloroform, CHCl3, is 61.700 °C at 1 atmosphere. Kb(chloroform) = 3.67 °C/m. In a laboratory experiment, students synthesized a new compound and found that when 10.55 grams of the compound were dissolved in 207.1 grams ofchloroform, the solution began to boil at 62.386 °C. The compound was also found to be nonvolatile and a non-electrolyte.

What is the molecular weight they determined for this compound ? g/mol

3.The freezing point of benzene, C6H6, is 5.500 °C at 1 atmosphere. Kf(benzene) = -5.12 °C/m. In a laboratory experiment, students synthesized a new compound and found that when 13.31 grams of the compound were dissolved in 269.1 grams of benzene, the solution began to freeze at 4.845 °C. The compound was also found to be nonvolatile and a non-electrolyte.

What is the molecular weight they determined for this compound ? g/mol

Explanation / Answer

Solution:- (1) From Raoult's law, "Relative lowering of vapor pressure is equal to the mole fraction of the solute.

(P0 - Ps)/P0 = Xsolute

Where, P0 = vapor pressure of pure solvent

Ps = vapor pressure of solution and

Xsolute = mole fraction of solute

Let's plug in the values in the formula...

(73.03 - 71.47)/73.03 = Xsolute

Xsolute = 0.0214

We know that, mole fraction of solute = moles of solute/(moles of solute + moles of solvent)

Let's calculate the moles of solvent that is benzene.

182.1 g benzene x (1mol/78.12 g) = 2.331 mol benzene

Let's say there are n moles of solute.

0.0214 = n/(n + 2.331)

on cross multiply.......

0.0214n + 0.0499 = n

0.0499 = n - 0.0214n

0.0499 = 0.9786n

n = 0.0499/0.9786

n = 0.05099

So, there are 0.05099 moles of the solute.

13.86 grams of it were used. So, molar mass = 13.86g/0.05099mol = 271.82 g/mol

(2) delta Tb = m*kb

where delta Tb = elevation in boiling point

m = molality of solution and

kb is the constant.

delta Tb = 62.386 - 61.700 = 0.686 degree C

kb = 3.67 degree C/m

So, m = 0.686/3.67 = 0.187

molality by definition is moles of solute per kg of solvent.

Solvent mass is given as 207.1 g that is 0.2071 kg.

0.2071 kg x (0.187 mol/kg) = 0.0387 mol

10.55 g of the compound were used, so, molar mass of the compound = 10.55g/0.0387mol = 272.61 g/mol

(3) delta Tf = m*kf

delta Tf = 4.845 - 5.500 = -0.655 degree C

kf = -5.12 degree C/m

m = (-0.655/-5.12) = 0.128

mass of solvent = 269.1 g = 0.2691 kg

So, 0.2691 kg x (0.128 mol/kg) = 0.0344 mol

molar mass of solute = 13.31g/0.0344 mol = 386.92 g/mol

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