Calculate the percent by mass of a solution containing 0.21 moles of c12h22o11 (gram-molecular mass = 342 g/mole) in 250 g of water.

28.728%

We are given;

Moles of the solute, C12H22O11 = 0.21 moles

Molar mass of C₁₂H₂₂O₁₁ = 342 g/mol

Mass of the solvent water = 250 g

We need to calculate the percent by mass of a solution

Mass = Number of moles × Molar mass

Therefore;

Mass of solute = 0.21 moles × 342 g/mol

                        = 71.82 g

% by mass = (Mass of solute ÷ mass of the solution)× 100

                = (71.82 g ÷ 250 g) × 100

                = 28.728%

Therefore, the percent by mass of the solution is 28.728%

thats a little if u want the answer just look it up on google thats what i do haha : )

what are the statements