# Question #b2f75

##### 1 Answer

#### 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 **every** milliliter of solution has a mass 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 **molality** 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

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")#