Question

At 300 K,36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
A. $11gL^{-1}$
B. $22gL^{-1}$
C. $36gL^{-1}$
D. $42gL^{-1}$

Answer 1

Hint: Use the formulae that define the relationship between the osmotic pressure and the molar concentration. This question is from the chapter solutions that have explained the different colligative properties of the compounds.

Complete step by step answer:
As per the van't Hoff equation, the relationship between the osmotic pressure and the molar concentration is Π=CRT. Here, the perfect gas constant is R, and the absolute temperature is T.
C1=36/180M
(Note: Molar mass of glucose is 180 g/mol and molar concentration is the ratio of several moles of glucose to the volume of solution in L. Number of moles is the ratio of mass to molar mass).
$\Pi_1=4.98 \mathrm{bar}$
$C_{2}=?$
$\Pi_2=1.52 \mathrm{bar}$
$\Rightarrow 4.98=\dfrac{180}{36 R T} \ldots \ldots(i)$
$\Rightarrow 1.52=C_{2} \mathrm{RT} \ldots \ldots(i i)$
On dividing equation 1 and 2 we get
$ \Rightarrow \dfrac{C_{2}}{36 \times 180}=4.981 .52$
$\Rightarrow C_{2}=0.061 \mathrm{M}$
Hence, the second solution has a concentration of 0.061 M.
Now multiply the equation with the molecular mass of the glucose that is 180.156 g/m
After the conversion, we get the following solution as
$=0.061 \times 180.156=10.98$
$\Rightarrow C_2 \,=10.98 \approx 11$

Therefore the correct option is option A.

Note:
Studying the colligative properties along with the determination of mass would be helpful in answering the questions of the same type. Also never forget to focus on the units given and the conversions that take place cause missing on a conversion can lead to a wrong answer.