Question

So $0.1M$ molar solution of urea has osmotic pressure of $400{\text{ mm Hg}}$ . 200 $200c{m^3}$ of $0.25M$ glucose and $300c{m^3}$ of $0.25M$ cane sugar are mixed. The osmotic pressure of this mixture would be?
(A) $2000{\text{ mm Hg}}$
(B) $1600{\text{ mm Hg}}$
(C) $80{\text{ mm Hg}}$
(D) $300{\text{ mm Hg}}$

Answer 1

Hint : Osmotic pressure is given by Osmotic pressure $(\pi ) = C \times R \times T$ , and hence, we need two values, the concentration and the temperature. From the first line of question, we can find out the temperature value and from the next line, that is the target, we can put the value of temperature and finally get the osmotic pressure.

Complete Step By Step Answer:
In order to solve this, we need to understand:
Osmotic pressure $(\pi ) = C \times R \times T$ , where, C= concentration(in moles/ltr); R= universal gas constant; T= temperature in kelvin.
First, we need to find the temperature of the system. It is given that 0.1 molar solution of urea has the osmotic pressure of 400 mm Hg. We know that:
$(\pi ) = C \times R \times T$
$400mmHg = 0.1M \times 62.36 \times T$
$T = 64.14K$
Next, we need to find the osmotic pressure of the mixture.
$(\pi ) = C \times R \times T$
Now, concentration= $\dfrac{{moles}}{{volume(in{\text{ ltr)}}}}$
Thus, concentration of the mixture= $\dfrac{{0.25 \times 0.2 + 0.3 \times 0.5}}{{0.5}} = \dfrac{{0.2}}{{0.5}} = 0.4$
Thus, the osmotic pressure of the solution$\left( \pi \right)$= $0.4 \times 62.36 \times 64.14$
$= 1600mmHg$
Hence, the osmotic pressure of the given solution mixture is $1600mmHg$ .
Thus the correct option is option(B).

Note :
it is not necessary that all the properties of the two given solutions (liquids) will add up. Here, since it is the properties of liquid, that is why osmotic pressure of the given individual liquid can be calculated by finding the concentration of the liquid mixture. Also, keep in mind that the value of universal gas constant changes when we change the units we want the answer in. Thus, whenever entering the equation values, make sure they have the same scale of units and if in question, the scale unit isn’t the same, we first have to have to convert the scale then have to make the necessary calculations and get to the final answer.