What mass of sucrose must be taken for solution weight of 200 g, which is equal to molality 0.05 mol / kg?
To determine the mass of sucrose needed for a solution with a weight of 200 g and a molality of 0.05 mol/kg, you need to use the formula for molality:
Molality (m) = moles of solute / mass of solvent (in kg)
Continue with the following steps to solve the problem:
Step 1: Convert the given weight of the solution from grams to kilograms.
200 g = 200 g / 1000 = 0.2 kg
Step 2: Rearrange the molality formula to solve for moles of solute.
moles of solute = molality * mass of solvent
moles of solute = 0.05 mol/kg * 0.2 kg
moles of solute = 0.01 mol
Step 3: Determine the molar mass of sucrose (C12H22O11) using the periodic table.
C = 12.01 g/mol
H = 1.008 g/mol
O = 16.00 g/mol
Molar mass of sucrose = (12.01 * 12) + (1.008 * 22) + (16.00 * 11)
Molar mass of sucrose = 144.12 + 22.176 + 176.00
Molar mass of sucrose = 342.296 g/mol
Step 4: Convert moles of solute to grams of solute using the molar mass of sucrose.
mass of sucrose = moles of solute * molar mass of sucrose
mass of sucrose = 0.01 mol * 342.296 g/mol
mass of sucrose = 3.42296 g
Therefore, you would need approximately 3.42 grams of sucrose for a solution with a weight of 200 g and a molality of 0.05 mol/kg.