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Calculate the mass of C12H22O11, contained in 500.0 mL of solution if the measured vapour pressure of the solution is 121.2 mbar at 50 °C.?
Calculate the mass of lactose, C12H22O11, contained in 500.0 mL of solution if the measured vapour pressure of the solution is 121.2 mbar at 50 °C. The vapour pressure of pure water at 50 °C is 123.3 mbar; assume no change in volume occurs upon dissolution.
Please help me with this. I get the mole ratio, but I don't know what to do from there. I think I'm supposed to use it to find the number of moles of lactose but I don't know how. Please help.
1 Answer
- Roger the MoleLv 74 years agoFavorite Answer
Let z be the mole fraction of C12H22O11. Then (1-z) is the mole fraction of water.
1-z = 121.2 mbar / 123.3 mbar = 0.982968
z = 1 - 0.982968 = 0.017032
Take a hypothetical sample of exactly 1.000 mol of solution:
(0.982968 mol H2O) x (18.01532 g H2O/mol) = 17.7085 g H2O
(0.017032 mol C12H22O11) x (342.2965 g C12H22O11/mol) = 5.82999 g C12H22O11
So the mass fraction of water is: (17.7085 g H2O) / (17.7085 g + 5.82999 g) = 0.75232
The problem states: assume no change in volume occurs upon dissolution. So if there was 500.0 mL of solution after dissolution, there must have been 500.0 mL of pure water before the dissolution. 500.0 mL of pure water would weigh 500.0 g.
(500.0 g H2O) / (0.75232) = 664.61 g solution
(664.61 g solution) - (500.0 g H2O) = 164.6 g C12H22O11