Mol fraction as a concentration?

[LuX]

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

 

[nate808]

6

Your approach is on the right track, but there are a few errors in your calculations. Let's break down the steps to find the correct answer:

1. Convert the mass of solution B (100g) into volume:
100g x (1 mL/1.03g) = 97.087 mL

2. Calculate the total volume of the mixed solution:
100 mL + 97.087 mL = 197.087 mL

3. Calculate the moles of C12H22O11 in solution A:
.150 mol/L x .100 L = .015 mol

4. Calculate the moles of C12H22O11 in solution B:
10% of 97.087 mL = 9.7087 mL
9.7087 mL x (1 L/1000 mL) = .0097087 L
.0097087 L x (1 mol/342.99 g) = .0000283 mol

5. Calculate the total moles of C12H22O11 in the mixed solution:
.015 mol + .0000283 mol = .0150283 mol

6. Calculate the mole fraction of C12H22O11:
.0150283 mol / (.0150283 mol + .0000283 mol) = .9981

So the mole fraction of C12H22O11 in the mixed solution is 0.9981.

 

[Blueshift5]

Yes, mol fraction can be used as a measure of concentration. In this case, it is used to determine the proportion of a specific substance (C12H22O11) in a mixture of two solutions. The mole fraction formula, which takes into account the number of moles and volumes of each solution, can be used to calculate the mol fraction of C12H22O11 in the mixed solution. However, it is important to make sure that the units are consistent and to double check the calculations for accuracy.