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LuX
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Homework Statement
2 solutions of C12H22O11 are mixed. Solution A is .150M with density of 1.05 g/mL. Solution B is 10%
C12H22O11 by mass with density of 1.03g/mL. 100 mL of solution A is mixed with 100g of B. What is the mole fraction of C12H22O11 in the mixed solution?
Homework Equations
mole fraction = (mol of A + mol of B)/ total mols and also (volume of A + volume of B)/total Volume
The Attempt at a Solution
i tried converting both solutions into volumes and using the mole fraction formula regarding volumes to calculate mol fraction. but i doubt i got it right...
for volume of solution A : .100L (.150mol/L)(342.99g/mol)(1 mL/1.05g) = 4.899 mL
for B: 10g(1mL/1.03g) = 9.708737
mol fraction = 4.899/14.60 = .335