-1
$\begingroup$

Work

mole fraction of NaCl = 0.21/0.21+0.79

If there is 1 mol of total solution, there will be 0.21 mol NaCl and 0.79 mol H2O

0.21 mol NaCl = 12.27 g

0.79 mol H2O = 14.23 g

this would yield a ratio of 12.27 : 14.23 inside the solution, giving 46.26 as the number of grams of NaCl inside 100 ml of solution.

Problem

I've looked around google for answers to this problem and many of them come up with 86.4 grams of NaCl in the solution. I feel like my answer is logical, but I'm not sure. Could you explain the mistakes I made in my work, if any?

$\endgroup$

closed as off-topic by Mithoron, A.K., Todd Minehardt, Jon Custer, Tyberius Feb 26 at 19:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ hint: Look up the density of water in g/mL $\endgroup$ – Night Writer Feb 24 at 19:49
0
$\begingroup$

The concentration you have calculated is % ($w/w$). In other words, what that means is mass of solute in $\pu{100 g}$ of solution. In your case, it is mass of $\ce{NaCl}$ in $\pu{100 g}$ of solution: $\frac{12.27}{(12.27+14.23)}\times 100=46.3$%. However, in your statement, you have declared it is % ($w/v$), which is not.

the others's answer is also not quite correct, because they have assumed when $\ce{NaCl}$ is dissolved in certain volume of water, the volume stays same (won't change). If you are allowed to assume that, then, you can conclude (like others) that when $\pu{12.27 g}$ of $\ce{NaCl}$ is dissolved in $\pu{14.23 g}$ of $\ce{H2O}$ (or $\pu{14.23 mL}$ of $\ce{H2O}$ assuming density of water is $\pu{1.00 g/mL}$), the volume of solution is still $\pu{14.23 mL}$. Based on allowed assumption, the % $(w/v)$ concentration of your solution is: $\frac{12.27}{14.23}\times 100=86.2$% (the value close enough to your internet finding).

However, most important data missing in your question is the density ($\rho$) of the sought $\ce{NaCl}$ solution. That would give exactly the volume ($v$) of the solution since $v=m/\rho$. Thus, correct % $(w/v)$ concentration of your solution is: $\left(\frac{12.27}{\rho(12.27+14.23)}\times 100\right)$%.

$\endgroup$
0
$\begingroup$

Problem:

An aqueous solution of NaCl has a mole fraction of 0.21. What is the mass of NaCl dissolved in 100.0 mL of solution?

Frankly this is a stupid problem as stated. First a solution with that much $\ce{NaCl}$ must have a density significantly greater than 1 g/ml. So there isn't any way to solve the problem with the information given. You need a density or something else. Second, I doubt that much $\ce{NaCl}$ can be dissolved in water.

In its Solubility table Wikipedia gives that the solubility of $\ce{NaCl}$ is 38.99 grams of $\ce{NaCl}$ per 100 grams of water. MW of $\ce{NaCl}$ is 58.443 g/mol and for water is 18.02 g/mol. So 38.99 grams of $\ce{NaCl}$ is 38.99/58.443 = 0.667 moles. A 100 grams of water is 100/18.02 = 5.549 moles.

0.667/(0.667+5.549) = 10.7%

So the greatest mole fraction that a sodium chloride solution can have is 10.7%.

$\endgroup$
-2
$\begingroup$

I cant understand your question but I will tell you some things that may be helpful to you .

Salts are fully dissolved in water so what you say that there will be a few moles of NaCl remaining is a mistake.Find the concetration of NaCl in your aqueous solution . I assume that your solution isn't dense so you must look for concetrations below 1 Molarity. 1 Molarity is 1 mole/ 1L = 1 mole/1000ml . Divide by 10 since you said that your solution has volume of 100mL and then you will find the moles. The total mass of NaCl in your solution is Molecular Mass of NaCl*Moles

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.