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# Calculate the number of moles of CH3OH in 5.00L of 2.00m solution if the density of the solution is .950 g/mL?

Here's what I got.

Your starting point here will be to convert the volume of the solution to mass by using the given .

So, you know that the solution has a of ##"0.950 g mL"^(-1)##, which is equivalent to saying that **every** milliliter of solution has a mass of ##"0.950 g"##.

In your case, the ##"5.00-L"## sample will have a mass of

##5.00 color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "0.950 g"/(1color(red)(cancel(color(black)("mL")))) = "4750 g"##

Now, this value represents the **total mass** of the solution, i.e. the mass of the , which is methanol, ##"CH"_3"OH"##, and the mass of the , which is water.

This will be your first equation

##m_"sol" = m_"W" + m_(CH_3OH)" " " "color(purple)((1))##

The solution is said to have a of ##"2.00 molal"##. As you know, a solution's is defined as the number of moles of solute divided by the mass of the solvent - expressed in **kilograms**.

##color(blue)(b = n_"solute"/m_"solvent")##

The problem is that you don't really know how much solvent you have in your sample. You know the solution's **total mass**, but not the mass of water.

As you know, the **molar mass** of a substance tells you the mass of **one mole** of that substance. This means that you can use the molar mass of methanol to find a relationship between the mass of the solute and the number of moles of solute present in the sample

##color(blue)(M_M = m/n)##

This means that you have

##m_(CH_3OH) = M_M * n" " " "color(purple)((2))##

Here ##n## represent the number of moles of methanol and ##M_M## is methanol's molar mass, which is equal to ##"32.042 g mol"^(-1)##.

Now use the solution's molality to find an equivalent expression for the mass of the solvent

##b = n/m_"W" implies m_"W" = n/b" " " "color(purple)((3))##

Plug equations ##color(purple)((2))## and ##color(purple)((3))## into equation ##color(purple)((1))## to get

##m_"sol" = overbrace(n/b)^(color(red)(M_"W")) + overbrace(M_M * n)^(color(brown)(m_(CH_3OH)))##

This will be equivalent to

##m_"sol" = n * (1/b + M_M)##

which will of course give you

##n = m_"sol"/(1/b + M_M)##

Now, it's **very important** to realize that molality is measured in moles per kilogram, so make sure to convert it to moles per gram first by using the conversion factor

##"1 kg" = 10^3"g"##

You will have

##"2.00 mol" color(red)(cancel(color(black)("kg"^(-1)))) * (1 color(red)(cancel(color(black)("kg"))))/(10^3"g") = 2.00 * 10^(-3)"mol g"^(-1)##

Now you're ready to plug in your values

##n = (4750 color(red)(cancel(color(black)("g"))))/(1/(2.00 * 10^(-3)) color(red)(cancel(color(black)("g"))) "mol"^(-1) + 32.042 color(red)(cancel(color(black)("g"))) "mol"^(-1))##

##n = "8.929 moles"##

Rounded to three , the answer will be

##n = color(green)("8.93 moles CH"_3"OH")##