Question

1.Calculate the number of moles of solute present in am aqueous solution of 86.3 g of 0.160 m KCl.

2. The density of toluene (C7H8) is 0.867g/mL, and the density of thiophene (C4H4S) is 1.065 g/mL. A solution is made by dissolving 9.38g of thiophene in 270 mL of toluene. Calculate the (a) percentage and (b) Molality of thiophene.

Answer 1

1.Calculate the number of moles of solute present in am aqueous solution of 86.3 g of 0.160 m KCl.

assume 0.16 mol of KCl, that is, mass of KCL = mol*MW = 0.16*74.5513 = 11.928 g of KCl

then,

1 kg of solvent is present;

total mass = 11.928 + 1000 = 1011.928 g

if we only have... 86.3 g of solution, of which it has 0.16 mol

86.3 g of solution = 0.16 mol of solute

1011.928 g of solution = x mol

x = 0.16*86.3/1011.928

x = 0.0136452 mol of KCl

2. The density of toluene (C7H8) is 0.867g/mL, and the density of thiophene (C4H4S) is 1.065 g/mL. A solution is made by dissolving 9.38g of thiophene in 270 mL of toluene. Calculate the (a) percentage and (b) Molality of thiophene.

% by mass:

Total mass of toluene = D*V = 270*0.867 = 234.09 g of toluene

mass of thiphene = 9.38 g

% of thiophene = mass of thiophene / total mass *100 = 9.38/(234.09+9.38) * 100 = 3.85263%

now, use M

M = mol of solute / kg of solvent

MW of tiophene

mol of thiophene = mass/MW = 9.38/84.14 = 0.1114808652 mol of tiophene

mas of solvent = 234.09 g = 0.23409 kg

molality = mol / kg solvent= 0.1114808652/0.23409 = 0.47623 molal