Question

can someone please help. A sucrose solution is prepared to a final concentration of 0.170M . Convert this value into terms of g/L, molality, and mass % (molecular weight, MWsucrose = 342.296g/mol ; density, ρsol′n = 1.02g/mL ; mass of water, mwat = 961.8g ). Note that the mass of solute is included in the density of the solution,

Answer 1

From Molarity: we have 0.17 mole sucrose in 1 Liter solution
1) To convert it into g/L we have to multiply moles of sucrose by its molar mass:
= 0.17 mole sucrose x
`(342.296 g sucrose)/(1 mole sucrose)` = 58.2 g/L
2) Molality = number of moles of solute / mass in kg of solvent =
`(0.17 mole solute)/(0.9618 kg solvent)` = 0.177 mole / kg
3) mass % = mass of solute / mass of solution x 100
mass of solution = 1020 g from its density
mass of solute = 1020 - 961.8 = 58.2 g solute
mass % =
`(58.2 g solute)/(1020 g solution)` x 100 = 5.7 %