Final answer:
To calculate the molality of a 36.0% aqueous HCl solution with a density of 1.20 g/mL, we assume a 100 g solution for ease of calculation. 36.0 g of HCl corresponds to 0.987 moles. Dividing this by 0.768 kg of water (given the solution density) provides a molality of 12.85 molal, with the closest provided option being (B) 12.00 molal.
Step-by-step explanation:
To calculate the molality of a 36.0% by mass aqueous solution of HCl, we need to follow several steps. Firstly, assume we have 100 g of the solution which will contain 36.0 g of HCl and 64.0 g of water. Using the molar mass of HCl (36.46 g/mol), we can convert the mass of HCl to moles:
36.0 g HCl × (1 mol HCl / 36.46 g HCl) = 0.987 moles of HCl
Since the density of the solution is 1.20 g/mL, the mass of 1 liter of the solution will be 1200 g. The mass of water in this liter of solution will therefore be:
1200 g × (100% - 36.0%) = 768 g of water
Molality is calculated by dividing the moles of HCl by the mass of water in kilograms:
(0.987 moles HCl) / (0.768 kg water) = 1.285 or 12.85 molal
However, the options provided do not include 12.85 molal, suggesting there may be a rounding or precision issue when handling the initial values or in the options provided. Nonetheless, based on the calculations here, the closest option is: (B) 12.00 molal