# Molarity to Percent, Mole Fraction, Molality

I have a series of questions similar to the following. I want to make sure I do this correct before I do the others. I am in need of a hint.

Given molarity of 6M NaOH with d=1.4g/mL, calculate, percent, mole fraction, and molality.


So in the above question, I have the molarity, but how am I supposed to get liters of solution?

I figured it was 6M per 1L of water, but then what am I supposed to do with the density? Is that the density of the water with the NaOH?

Then in a sub part it gives me the mole fraction as 0.1 with a density of 1.3g/mL. Why does the question give me that number? Is it possible to figure out percent, molality, and molarity based on mole fraction?

Thanks for you help.

I figured it was 6M per 1L of water

That is incorrect you have 6 mol per 1L of water. M(stands for mol/L) not for mol!

How to tackle this problem? Check water density, is 1 $g/cm^{2}$. So yours is the density of the solution. Then understand what the problem is asking what is percent? What is mole fraction? What is molality? Are you able to answer these questions? If not go back to the text book. Then is only a matter of simple algebra and units matching. And remember if you have weight and molecular weight you can calculate the number of mole!

So suppose you have a liter of this solution what's the weight of NaOH?

molecular weight ($\frac{g}{mol}$) of NaOH $\times$ number of moles

Now what's the Overall weight?

Overall weight=density $\times$ volume (remember to use the right units)

Percent what? In chemistry there are a lot of percent, you mean percentage by weight (wt%) I imagine. Try to figure out from your self:

$= \frac{NaOH weight}{Overall weight}\times100$

Mole fraction require mole of both, we start from the weight, so whats the weight of water?

you got the overall weight and the weight of NaOH

Now calculate the moles of water, and it is simple to find the mole fraction, and molality. Make these steps in the opposite way and you will find the answer to your last question.