500g urea aq. Sol. Has mole fraction 0.2 what is molality of sol.
3 answers
answer to how much heat is required to raise the temperature of 68.0 g of AIF3 from 25 degrees celsius to 80.0 degrees celsius
Your question is answered above.
The mass of the solution is 500. How much is urea and how much is water?
Let X = g urea
and 500-x = grams H2O
Then mols urea = X/60
mol H2O = (500-X/18)
Then
nurea/(nurea + nH2O) = 0.2
(X/60)/{[(X/60)] + [(500-X)/18]} = 0.2
I solved for X and obtained about 227 g for urea and 273 g H2O. You can work that out but that's very close to a mole fraction for urea of 0.2.
Then n urea = about 3.8
n H2O = about 15.2
From above we had 227 g urea (about 3.8 mols) and 273 g H2O
molality = (mols solute)/(kg solvent)
mols solute = about 3.8
kg solvent = about 0.273
m = ?
My numbers are estimates; you should recalculate each step.
Let X = g urea
and 500-x = grams H2O
Then mols urea = X/60
mol H2O = (500-X/18)
Then
nurea/(nurea + nH2O) = 0.2
(X/60)/{[(X/60)] + [(500-X)/18]} = 0.2
I solved for X and obtained about 227 g for urea and 273 g H2O. You can work that out but that's very close to a mole fraction for urea of 0.2.
Then n urea = about 3.8
n H2O = about 15.2
From above we had 227 g urea (about 3.8 mols) and 273 g H2O
molality = (mols solute)/(kg solvent)
mols solute = about 3.8
kg solvent = about 0.273
m = ?
My numbers are estimates; you should recalculate each step.
2 answers