What mass of sucrose must be taken for solution weight of 200 g, which is equal to molality 0.05 mol / kg?
To find the mass of sucrose needed for a solution with a given weight and molality, you'll need to use the formula:
Molality (m) = moles of solute / mass of solvent (in kg)
Rearranging the formula, we get:
moles of solute = Molality (m) * mass of solvent (in kg)
In this case, the molality is given as 0.05 mol/kg, and the mass of the solvent is 200 g (which is 0.2 kg since there are 1000 grams in a kilogram).
Substituting the values into the formula:
moles of solute = 0.05 mol/kg * 0.2 kg
moles of solute = 0.01 mol
Sucrose (C12H22O11) has a molar mass of 342.3 g/mol. So to find the mass of sucrose needed, we multiply the moles of sucrose by its molar mass:
mass of sucrose = moles of sucrose * molar mass of sucrose
mass of sucrose = 0.01 mol * 342.3 g/mol
mass of sucrose = 3.423 g
Therefore, the mass of sucrose required for a solution with a weight of 200 g and molality of 0.05 mol/kg is 3.423 grams.