$\pu{1060 g}$ of urea aqueous solution is subjected to heating. It starts boiling at $\pu{100.53 °C}.$ It is heated up to $\pu{102.12 °C}.$ Find amount of water vapourised in grams. (Pure water boiling point is $\pu{100 °C},$ $K_\mathrm{b}$ of $\ce{H2O}$ is $\pu{0.53 K kg mol-1}.)$

When water evaporates will the equivalent amount of urea also evaporate i.e. will we calculate final molality with the same amount of solute or a decreased amount?

I first calculated the initial amount of water as (53000/3) g and then taking the same moles of urea calculated the final amount which came out to be (53000/12) g. So the amount of water evaporated should be 13250 g but the given answer is 750 g.

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    $\begingroup$ please elaborate your doubt here, did you try the question yourself? Do you know the concept? Where are you stuck? $\endgroup$ – Haha Hahaha Jan 2 '20 at 6:37
  • $\begingroup$ We have a policy which states that ‎you should show your thoughts, effort and attempts to answer your question yourself. It'll make us certain that ‎we aren't doing your homework for you, and that the Q/A is beneficial for broad audience. As "homework-like questions" are considered literal homeworks, self-study questions, puzzles, worked examples etc. Please edit in your full reasoning or thoughts on this. See Homework. Otherwise, the question may get closed as "low OP effort homework-like question." $\endgroup$ – Poutnik Jan 2 '20 at 7:09
  • $\begingroup$ Hint: when water (only) will vaporise, molality ↑ and hence ∆Tb ↑ $\endgroup$ – Rahul Verma Jan 2 '20 at 7:16
  • $\begingroup$ It's 13250g only. This time mypat answer key was completely flawed(except for maths). Most probably someone hacked the system as someone got 300 with this totally flawed key $\endgroup$ – user600016 Jan 2 '20 at 13:24
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    $\begingroup$ I am not obviously not talking about that data to be given in the question, as @Rahul Verma said, you are given 1 molal urea solution, and if you do the math you can find out that there's one mole of urea given and the rest of 1kg (1060g-(60*1)g) is water. $\endgroup$ – Sir Arthur7 Jan 3 '20 at 16:01

While considering colligative properties we always assume solute to be non-volatile. So, it doesn't contribute to the vapor. The amount of vapor will only be due to solvent particles. So, in the problem we can use the following relation:
$$ \Delta T = K_\mathrm b\cdot m $$ Initially $ \Delta T = 0.53\ \mathrm{^\circ C}$ so plugin the value of $K_\mathrm b$ and you end up with the value of initial molality of solvent as $1$. From this you can find the initial moles of solute which remain constant throughout.

Next you put $ \Delta T = 2.12\ \mathrm{^\circ C}$ and plug in the known values of $K_\mathrm b$ and 'moles of solute' in the above formula to get the final value of solvent amount that remains in the solution. Subtracting the final amount of solvent left from the initial amount will let you know the amount of water vaporized which is $750\ \mathrm g$.

  • $\begingroup$ That solution gives 13.25 Kg. Just tried it again $\endgroup$ – Pheonix MC Jan 3 '20 at 12:43

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