Question

What is the amount of urea dissolved per litre, if aqueous solution is isotonic with 10% cane sugar solution: (mol. Wt of urea=60)?
A.$200g/L$
B.$19.2g/L$
C.$17.54g/L$
D.$16.7g/L$

Answer 1

Hint: Isotonic solutions are those solutions which have the same osmotic pressure. Try to find a relation by using the data of the solutions given in the question, substitute them and obtain the weight.

Complete answer:
In order to answer our question, we need to learn about isotonic solutions. Two solutions that have the value of pressure in both are called isotonic solutions. The isotonic solutions at the same temperature have the same molar concentration. If we have solutions having different osmotic pressures, then the solution which has comparatively more osmotic pressure is called hypertonic solution and the other one which possesses low pressure is called hypotonic solution.$0.9%$(mass by volume) aqueous solution of sodium chloride is called normal saline solution which is isotonic with the fluid inside the human red blood cells(RBC). Therefore, medicines are mixed with saline water before the injections. This prevents the blood cells from shrinking or bursting. If the solute is less concentrated than water will move into the cell and it will swell up and burst. On the other hand, if it is concentrated, then water will move out of the cell and the cell will shrink, causing plasmolysis. Mathematically, the relation is:
     \[\begin{align}
  & \pi V=nRT \\
 & \pi =\dfrac{n}{V}RT \\
\end{align}\]
Here,$\pi $ is the osmotic pressure, V volume, T temperature and R universal gas constant. Now, let us come to our question. We can assume the solution to be 100mL out of which 10mL is cane sugar, having molar mass of $342g\,mo{{l}^{-1}}$. We can write from the second equation that:
     \[\dfrac{{{W}_{urea}}}{mol.w{{t}_{urea}}\times volum{{e}_{urea}}}=\dfrac{{{W}_{sugar}}}{mol.w{{t}_{sugar}}\times volum{{e}_{sugar}}}\]
Now, by substituting the values given in the question, we can see that:
     \[\begin{align}
  & \dfrac{{{W}_{urea}}}{60\times 1}=\dfrac{10}{342\times 0.1} \\
 & {{W}_{urea}}=\dfrac{60\times 10}{342\times 0.1}=17.54g/L \\
\end{align}\]

So, we obtain our answer as $17.54g/L$ , which comes out to be option C.

Note:
Common examples of osmosis include:
A. Shrivelling of raw mangoes to pickle
B. Swelling of tissues in people consuming more salt
C. Preservation of meat
D. Absorption of water by plants