Question #31579

Concentrated Hydrochloric acid is commonly sold as 12.0 M HCL and its density is 1.18 g/cm^3 . Calculate: (a)the Molality of the solution. (b) the weight percent of HCL in the solution

Expert's answer

The molality, b, of a solution is the amount of substance (in mol) of solute, divided by the mass (in kg) of the solvent (not the mass of the solution):

b = n/m

1 L of the HCl solution contains 12 mol of the HCl. The mass of HCl is:

m(HCl) = n(HCl)*M(HCl) = 12 mol * 36.5 g/mol = 438 g=0.438 kg

The mass of the solution is:

m(solution) = V(solution) * d(solution) = 1 L * 1.18 kg/L = 1.180 kg

The mass of the solvent (water) is:

m(H2O) = m(solution) - m(HCl) = 1.180 - 0.438 = 0.742 kg

a)

The molality of the solution is:

b(solution) = 12 mol/0.742kg = 16.17 m HCl

b)

The weight percent of HCL in the solution can be calculated according to the following equation:

w(solute) = m(solute)/m(solution)

w(HCl),% = (0.438 kg / 1.180 kg)*100% = 37.12%

b = n/m

1 L of the HCl solution contains 12 mol of the HCl. The mass of HCl is:

m(HCl) = n(HCl)*M(HCl) = 12 mol * 36.5 g/mol = 438 g=0.438 kg

The mass of the solution is:

m(solution) = V(solution) * d(solution) = 1 L * 1.18 kg/L = 1.180 kg

The mass of the solvent (water) is:

m(H2O) = m(solution) - m(HCl) = 1.180 - 0.438 = 0.742 kg

a)

The molality of the solution is:

b(solution) = 12 mol/0.742kg = 16.17 m HCl

b)

The weight percent of HCL in the solution can be calculated according to the following equation:

w(solute) = m(solute)/m(solution)

w(HCl),% = (0.438 kg / 1.180 kg)*100% = 37.12%

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