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
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Expert's answer
2013-06-10T11:03:12-0400
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%
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