Question

.

What is the HCl concentration (molality) of the solution prepared by mixing 0.100 L of 4.729 M HCl that has a density of 1.0776 g mL ^-1 with 56.0 g of an HCl solution with a mole fraction of 0.1070?

Answer 1

0.100 L of 4.729 M HCl = volume of HCl solution in L * molarity = 0.100 L * 4.729 mole / L = 0.4729 mole.

volume of solution = 0.100 L = 100 ml

mass of HCl = 0.4729 mole * 36.5 g / mole = 17.26 g

mass of the solution = density * volume = 1.0776 g / ml * 100 ml = 107.76 g

mass of water = (107.76 - 17.26) = 90.5 g

56.0 g of an HCl = mass / molar mass = 56.0 g / 36.5 g / mole = 1.534 mole.

mole fraction = 0.1070

or

moles of HCl / (mole of HCl + moles of water) = 0.1070

or

1.534 / (1.534 + mole of water) = 0.1070

or

mole of water = 12.8 mole.

mass of water = 12.8 mole * 18.0 g / mole = 230.49 g

thus

total mass of water solvent = (230.49 +90.5) = 321 g = 0.321 kg

total mole of HCl in mixture = (0.4729 + 1.534) = 2.0069 mole.

molality = 2.0069 mole / 0.321 kg = 6.252 mole / kg