Question

Two beakers A and B are present in a closed vessel. Beaker A contains 152.4g aqueous solution of urea, containing12 g of urea. Beaker B contains 196.2 g glucose solution, containing 18g of glucose. Both the solutions allowed us to attain equilibrium. Determine wt. % of glucose in its solution at equilibrium:
A. 25%
B. 10%
C. 18%
D. 14.49%

Answer 1

Hint: We know that solutions described in concentration like molarity, normality or molality. Solutions are sometimes represented in terms of relative percent concentration of solute in a solution. To determine the weight percent of a solution, divide the mass of solute by mass of the solution (solute and solvent together) and multiply by 100 to obtain per cent.

Complete step-by-step answer:
To answer this question, we should calculate mole fraction of both the compounds that are present in solution.
Beaker A:
So, first we should calculate mole fraction of urea.
\[Mole\,fraction\,of\,solute=\dfrac{Moles\,of\,solute}{Moles\,of\,solute+Moles\,of\,sovent\,}\]
\[Mole\,fraction\,of\,urea=\dfrac{\dfrac{12}{60}}{\dfrac{12}{60}+\dfrac{140.4}{18}\,}=\dfrac{0.2}{0.2+7.8}=0.025\]
Beaker B:
\[Mole\text{ }fraction\text{ }of\text{ }glucose=\dfrac{\dfrac{18}{80}}{\dfrac{18}{80}+\dfrac{178.2}{18}}=0.01\]
From the calculation, we observe that mole fraction of glucose is less so vapour pressure above the glucose solution will be higher than the pressure above urea solution, so some \[{{H}_{2}}O\]molecules will transfer from glucose to urea side in order to make the solutions of equal mole fraction to attain equilibrium. Let x mole of \[{{H}_{2}}O\] be transferred.
$\dfrac{0.2}{0.2+7.8+x}=\dfrac{0.1}{0.1+9.9-x}\to x=4$

Now, mass of glucose solution will be \[=196.2-18\times 4=124.2\]
$weight\%\,of\,glu\cos e\,will\,be=\dfrac{18}{124.2}\times 100\to 14.49\%$
So, from the above calculation, we can now say that weight% of glucose is 14.49% and option D is correct.

Note: We should know that concentrations are often expressed in terms of relative unites with different types of percentage concentrations commonly used:
Mass Percent: The mass percent is used to express the concentration of a solution when the mass of a solute and the mass of a solution is given:
\[Mass\text{ }Percent=\dfrac{Mass\text{ }of\text{ }Solute}{Mass\text{ }of\text{ }Solution}\times 100\%\] \[Mass\text{ }Percent=\dfrac{Mass\text{ }of\text{ }Solute}{Mass\text{ }of\text{ }Solution}\times 100\%\]
Volume Percent: The volume percent is used to express the concentration of a solution when the volume of a solute and the volume of a solution is given:
\[Volume\text{ }Percent=\dfrac{Volume\text{ }of\text{ }Solute}{Volume\text{ }of\text{ }Solution}\times 100\%\]