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ARIGATÕ
I WILL TRY AND GET A FEW THINGS TO BE DONE
Data:
<span>Solute: 28.5 g of glycerin (C3H8O3)
Solvent: 135 g of water at 343 k.
Vapor pressure of water at 343 k: 233.7 torr.
Quesiton: Vapor pressure of water
Solution:
Raoult's Law: </span><span><span>The vapour
pressure of a solution of a non-volatile solute is equal to the vapour
pressure of the pure solvent at that temperature multiplied by its mole
fraction.
Formula: p = Xsolvent * P pure solvent
X solvent = moles solvent / moles of solution
molar mass of H2O = 2*1.0g/mol + 16.0 g/mol = 18.0 g/mol
moles of solvent = 135 g of water / 18.0 g/mol = 7.50 mol
molar mass of C3H8O3 = 3*12.0 g/mol + 8*1 g/mol + 3*16g/mol = 92 g/mol
moles of solute = 28.5 g / 92.0 g/mol = 0.310 mol
moles of solution = moles of solute + moles of solvent = 7.50mol + 0.310mol = 7.810 mol
Xsolvent = 7.50mol / 7.81mol = 0.960
p = 233.7 torr * 0.960 = 224.4 torr
Answer: 224.4 torr
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