Step-by-step explanation:
To make water boil at 101.3°C, we need to increase its boiling point from 100°C to 101.3°C. This means we need to increase the boiling point elevation by ΔTb = 1.3°C.
The boiling point elevation is given by the formula:
ΔTb = Kb * m
where Kb is the molal boiling point constant for water (0.512°C/m) and m is the molality of the solution.
We can rearrange this formula to solve for m:
m = ΔTb / Kb
m = 1.3°C / 0.512°C/m
m = 2.54 m
This tells us that the molality of the solution must be 2.54 m in order to increase the boiling point of water by 1.3°C.
The molality of a solution is defined as the number of moles of solute per kilogram of solvent. We can use the molecular weight of sucrose to convert from mass to moles:
Molecular weight of sucrose = 12*12 + 22*1 + 11*16 = 342 g/mol
To find the mass of sucrose needed to make a 2.54 m solution in 3200 g of water, we can use the following equation:
molality = moles of solute / (mass of solvent in kg)
Rearranging this equation gives:
moles of solute = molality * mass of solvent in kg
moles of sucrose = 2.54 mol/kg * (3200 g / 1000 g/kg) = 8.128 mol
mass of sucrose = moles of sucrose * molecular weight of sucrose
mass of sucrose = 8.128 mol * 342 g/mol = 2775.936 g
Therefore, we need 2775.936 g (or about 2.78 kg) of sucrose to add to 3200 g of water to create a 2.54 m solution that will boil at 101.3°C.