Final answer:
The molality of 150.0 grams of barium bromide dissolved in 1200 mL of water is calculated to be 0.421 mol/kg, which does not match any of the provided answer choices.
Step-by-step explanation:
To calculate the molality of 150.0 grams of barium bromide (BaBr2) dissolved in 1200 mL of water, we first need to find the number of moles of BaBr2 and the mass of the water in kilograms.
Firstly, we calculate the molar mass of BaBr2:
- Barium (Ba) has an atomic mass of 137.3 g/mol.
- Bromine (Br) has an atomic mass of 79.9 g/mol, since there are two bromines in BaBr2 we multiply this by 2.
Molar mass of BaBr2 = 137.3 g/mol + (2 × 79.9 g/mol) = 297.1 g/mol.
Next, we find the number of moles of BaBr2 by dividing the mass by the molar mass:
Number of moles = 150.0 g ÷ 297.1 g/mol = 0.505 moles
We convert the mass of water from milliliters to kilograms (note that 1 mL of water has a mass of about 1g):
Mass of water = 1200 mL × 1 g/mL = 1200 g = 1.2 kg
Next, calculate the molality of the solution:
Molality (m) = number of moles of solute ÷ mass of solvent (kg)
Molality (m) = 0.505 moles ÷ 1.2 kg = 0.421 mol/kg
None of the given choices (a) 1.25, (b) 0.75, (c) 1.50, and (d) 2.00 mol/kg match the calculated molality. Therefore, the correct molality is not listed among the options provided in the question.