Question #f5a94

1 Answer
Jan 17, 2017

Answer:

Here's what I got.

Explanation:

I'll assume that the problem gave you #2.3%# as the percent concentration by mass of ethanol in water, meaning that the percent concentration by mass of water is

#"% H"_2"O" = 100% - 2.3% = 97.7%#

So, a solution's percent concentration by mass, #"% m/m"#, sometimes called percent concentration by weight, is calculated by looking at the number of grams of solute present in #"100 g"# of solution.

In your case, a #"2.3% m/m"# ethanol solution contains #"2.3 g"# of ethanol, your solute, for ever #"100 g"# of solution.

Now, molality, #b#, is defined as the number of moles of solute present for every #"1 kg"# of solvent.

If you take a sample of this solution that has a total mass of #"100 g"#, you will find that it contains

#color(blue)(cancel(color(black)(100))) color(red)(cancel(color(black)("g solution"))) * "2.3 g ethanol"/(100color(red)(cancel(color(black)("g solution")))) = "2.3 g ethanol"#

and

#color(blue)(cancel(color(black)(100))) color(red)(cancel(color(black)("g solution"))) * ("97.7 g H"_2"O")/(100color(red)(cancel(color(black)("g solution")))) = "97.7 g H"_2"O"#

Notice that because the solution contains only ethanol and water, you can find the mass of water by doing

#m_"water" = "100 g" - "2.3 g" = "97.7 g"#

Use the molar mass of ethanol to convert the grams to moles

#2.3 color(red)(cancel(color(black)("g"))) * "1 mole ethanol"/(46color(red)(cancel(color(black)("g")))) = "0.050 moles ethanol"#

Since you know that

#color(blue)(ul(color(black)("1 kg" = 10^3"g")))#

you can convert the mass of water to kilograms

#97.7 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.977 kg"#

This means that the solution's molality will be

#b = "0.050 moles ethanol"/"0.977 kg water" = color(darkgreen)(ul(color(black)("0.051 mol kg"^(-1))))#

The answer is rounded to two sig figs.