Mass percent chemistry problem, stuck.?

Question

What is the mass percent of sucrose (C12H22O11, Mm=342 g/mol) in a 0.173-m sucrose solution?

I know that mass percent=mass solute/mass solution x 100, but here they give us molality?

Answer 1

You need to use the equation molality = moles of solute / kilograms of solvent

To start this question, assume you have 1 kg of solvent so that your 0.173 m sucrose solution really means 0.173 moles of sucrose in 1kg of solvent.

Work out the mass of sucrose that you have in this 0.173 moles using mass = moles x molar mass. So mass = 0.173 mol x 342 g/mol = 59.17 g sucrose in the solution

Now you know your solvent is 1 kg (1000g) so all together, your solution is the 1000g of solvent PLUS your 59.17 g of sucrose = 1059.17g

Now work out the mass percent. Mass percent = 59.17g (mass solute) / 1059.17g (mass of solution)  x 100% = 5.59

Answer 2

by the molality you discover how much there is of mass in the solution

0,173mol -------- x

1 mol ------------- 342g

then you just do the math ;)

Answer 3

You are screwed.