Question

An aqueous solution is labeled as 0.25 m NaCl. What is the mole fraction of the solute in this solution? Show all calculations leading to an answer.

Answer 1

@iGreen

Answer 2

@StudyGurl14

Answer 3

Hold on, I have to check something.

Answer 4

I don't know....and I should know... @surry99

Answer 5

That's okay, could you try helping me with other questions? @StudyGurl14

Answer 6

i'll try

Answer 7

i have to eat breakfast soon t ho

Answer 8

An aqueous solution freezes at -3.47⁰C. What is its boiling point? Show all calculations leading to an answer. @StudyGurl14

Answer 9

@radar @uri @undeadknight26

Answer 10

@Somy

Answer 11

Is that a lower case m? For molality? If yes, 0.25 m means that there are 0.25 mol of NaCl for every kilogram of water.

To find the mole fraction, we need to know how many moles there are of NaCl and water. We already know moles of NaCl (it's 0.25), so we just need to know how many moles of water that corresponds to.

1 kg of water contains 1000 g / 18 g/mol = 55.555 mol.

So, in total there are 0.25+55.555=55.8 moles of stuff per "unit" of solution. Of this, 0.25/55.8 *100 = 0.45% is the mole fraction of NaCl (the solute).

Answer 12

As for your second question, now we're talking about colligative properties. The freezing point depression equation is:

\[\Delta T_f=iK_fm\]
Where ∆T(f) is the change in freezing point, i is the van't Hoff factor, K(f) is the "freezing point depression" constant, and m is the molality.

We're given ∆T(f) in the question, and I guess you need to know that for water, K(f) is 1.853 K kg/mol. We can combine i and m into one variable because their values will be the same for when we consider boiling point elevation (since you're using the exact same solution). So, solving for im we get:

\[3.47=i \times 1.853 \times m\]\[im=1.87\]

Now we use this to find the boiling point elevation. Again, I'm guessing you need to know that the "boiling point elevation" constant for water is 0.512 K kg/mol. This time we're solving for the temperature change:

\[\Delta T_b=iK_bm\]\[\Delta T_b=1.87 \times 0.512\]\[\Delta T_b=0.96\]
This means the boiling point of your solution has increased by 0.96 K (or degrees C). The boiling point will therefore be 100.96 degrees C (remember, water alone boils at 100 degrees C).

If you have any questions about either of my answers let me know!