What is the molality of a solution of sucrose(molar mass=342) dissolved in water that is 25% sucrose by mass?
Take 1000g of it, sugar is 250g, water is 750g
m=250/343)/.750 moles/kgsolvent
To calculate the molality (m), we need to know the mass of the solute (sucrose) and the mass of the solvent (water). Given that the solution is 25% sucrose by mass, we can assume we have 100 grams of the solution. Thus, we have 25 grams of sucrose (solute) and 75 grams of water (solvent).
To calculate the molality, use the formula:
m = moles of solute / mass of solvent (in kg)
First, we need to convert the mass of sucrose from grams to moles using the molar mass of sucrose, which is 342 g/mol:
moles of sucrose = mass of sucrose / molar mass of sucrose
moles of sucrose = 25 g / 342 g/mol ≈ 0.073 moles
Next, we need to convert the mass of water from grams to kilograms:
mass of water = 75 g / 1000 g/kg = 0.075 kg
Now we can calculate the molality:
m = 0.073 moles / 0.075 kg ≈ 0.97 mol/kg
Therefore, the molality of the sucrose solution is approximately 0.97 mol/kg.