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”)##

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