Question #8c3ef
1 Answer
Here's what I got.
Explanation:
Assuming that you're working with a mass by volume percent concentration,
#10 color(red)(cancel(color(black)("mL solution"))) * "5 g urea"/(100color(red)(cancel(color(black)("mL solution")))) = "0.5 g urea"#
Now, in order to find the solution's osmolarity, you need to find its molarity, i.e. the number of moles of solute present for every
To do that, calculate the number of grams of urea present in
#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.5 g urea"/(10color(red)(cancel(color(black)("mL solution")))) = "50 g urea"#
Now use the molar mass of urea to convert this to moles
#50 color(red)(cancel(color(black)("g"))) * "1 mole urea"/(60.06color(red)(cancel(color(black)("g")))) = "0.8325 moles urea"#
Since this represents the number of moles of urea present in
#"molarity = 0.8325 mol L"^(-1)#
To find the osmolarity of the solution, which tells you the number of osmoles of solute present in
As you know, an Osmole is simply a mole of particles that contribute to a solution's osmotic pressure. In this case, the fact that urea is a non-electrolyte implies that it does not dissociate in aqueous solution to produce ions.
Consequently, you can say that
Therefore, the osmolarity of the solution will be equal to
#0.8325color(white)(.) color(red)(cancel(color(black)("moles")))/"1 L solution" * "1 Osmole"/(1color(red)(cancel(color(black)("mole")))) = color(darkgreen)(ul(color(black)("0.83 Osmol L"^(-1))))#
I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for your values.