Question #f5a94
1 Answer
Here's what I got.
Explanation:
I'll assume that the problem gave you
#"% H"_2"O" = 100% - 2.3% = 97.7%#
So, a solution's percent concentration by mass,
In your case, a
Now, molality,
If you take a sample of this solution that has a total mass of
#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.