Question #b2f75
1 Answer
Here's what I got.
Explanation:
Your starting point here will be to convert the volume of the solution to mass by using the given density.
So, you know that the solution has a density of
In your case, the
#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 solute, which is methanol,
This will be your first equation
#m_"sol" = m_"W" + m_(CH_3OH)" " " "color(purple)((1))#
The solution is said to have a molality of
#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
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
#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 sig figs, the answer will be
#n = color(green)("8.93 moles CH"_3"OH")#