What is the molality of the solution when you have
48.2 % by mass KBr aqueous solution ?
Show each conversion step.
To determine the molality of a solution, we need to know the mass of the solute (KBr) and the mass of the solvent (water). In this case, we have a 48.2% by mass KBr aqueous solution, which means that 48.2 grams of KBr are dissolved in 100 grams of the solution.
To find the molality, we can follow these steps:
Step 1: Calculate the mass of KBr in the solution.
In a 100g solution, 48.2% of it is KBr. So, the mass of KBr = (48.2/100) * 100g = 48.2g.
Step 2: Calculate the mass of water in the solution.
Since the remaining 51.8% by mass is water, the mass of water = (51.8/100) * 100g = 51.8g.
Step 3: Find the moles of KBr.
To convert the mass of KBr to moles, we need to know its molar mass. The molar mass of KBr is the sum of the atomic masses of potassium (39.10 g/mol) and bromine (79.90 g/mol). So, the molar mass of KBr is 39.10 g/mol + 79.90 g/mol = 119.00 g/mol.
The number of moles of KBr = mass of KBr / molar mass of KBr = 48.2g / 119.00 g/mol ≈ 0.405 moles.
Step 4: Calculate the molality.
Molality (m) is defined as the moles of solute per kilogram of solvent. Since we have the mass of water, we need to convert it to kilograms (1 kg = 1000 g).
Mass of water in kg = 51.8g / 1000 = 0.0518 kg.
Molality (m) = moles of solute (KBr) / mass of solvent (water in kg).
Molality (m) = 0.405 moles / 0.0518 kg ≈ 7.82 mol/kg.
Therefore, the molality of the solution with a 48.2% by mass KBr aqueous solution is approximately 7.82 mol/kg.