Rault's Law, Finding Mass of an unknown.

[dolpho]

Homework Statement

When 10.7 g of an unknown, non-volatile, non-electrolyte, X was dissolved in 100. g of water, the vapor pressure of the solvent decreased from 23.8 torr to 23.6 torr at 298 K. Calculate the molar mass of the solute, X.

Homework Equations

The answer is 227, but I know I'm doing wrong in the equation somewhere.

The Attempt at a Solution

100gH20 = 5.56 moles, Molar Fraction = .991

.991 = 5.56 / 5.56 + x =

5.51 + .991x = 5.56

.05 / .991 = 05 <---subtracted

10.7 / .05 = 214 g

Would appreciate any help!

 

[Borek]

Don't round down intermediate values and you will be OK.

 

[dolpho]

Don't round down intermediate values and you will be OK.

Sigh I just got 227.7, I would have definitely rounded off but at least I got the answer now lol. Generally speaking, should I ever round off during the calculations? Some teachers do and some don't so I'm a little confused.

 

[Borek]

Never round intermediate values during calculations (you have just seen why). If you have to report them, report rounded, but in calculations either use all digits you got (if they are kept in calculator register), or - if you have to save some value to use it later, use at least 2 or 3 so called guard digits - that is, if you know that the number you calculated has 3 sig figs, write it down with 5 or 6 digits.

 

[DeeNos]

I would like to point out that the given information is related to Rault's Law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. Therefore, the equation used in the attempt at a solution is incorrect.

To correctly solve this problem, we can use the following equation:

P1/P2 = X2/X1

Where P1 is the initial vapor pressure of the solvent (23.8 torr), P2 is the final vapor pressure of the solvent (23.6 torr), X1 is the mole fraction of the solvent (calculated to be 0.991), and X2 is the mole fraction of the solute (unknown).

Rearranging the equation, we get:

X2 = (P2/P1) * X1

Substituting the values, we get:

X2 = (23.6/23.8) * 0.991 = 0.986

Now, we can use the definition of mole fraction to calculate the moles of the solute (X) present in the solution:

Mole fraction = moles of solute / (moles of solvent + moles of solute)

0.986 = moles of X / (5.56 + moles of X)

Solving for moles of X, we get:

moles of X = 5.51

Finally, we can use the definition of molar mass to calculate the molar mass of the solute:

Molar mass = mass of solute / moles of solute

Molar mass = 10.7 g / 5.51 moles = 1.94 g/mol

Therefore, the molar mass of the unknown solute is 1.94 g/mol. I hope this helps!