
Which of the following aqueous solutions are isotonic? ($R=0.082atm{{K}^{-1}}mo{{l}^{1}}$)
(This question has multiple correct options)
(A) $0.01M$ glucose
(B) $0.01M$ $NaN{{O}_{3}}$
(C) $500ml$ solution containing $0.3g$ urea
(D) $0.04N$$HCl$
Answer
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Hint: If the concentration of the solute on both sides of the semipermeable membrane is equal, then it is an isotonic solution. When the two solutions across a semipermeable membrane have the same osmotic pressure, then the solutions are called isotonic solutions.
Formula Used: The formula used to calculate the osmotic pressure of a solution is:
$\pi = icRT$; where $i$ = van’t hoff index
$c$ = Concentration of solute, $R$ = Gas constant, $T$ = Temperature (Kelvin), $\pi $= Osmotic pressure.
Complete Step by Step Answer:
For two solutions to be isotonic, they should have the same osmotic pressure. So, we will calculate the osmotic pressure of each of the solutions given in the options, and the two solutions that will have the same osmotic pressure will be our answer.
(A) $0.01M$ glucose
As glucose does not dissociates, that is, it does not give any ions, so $i=1$
Thus, osmotic pressure will be
$\pi =1\times 0.01\times R\times T$
$\pi =0.01RT$
(B) $0.01M$ $NaN{{O}_{3}}$
As $NaN{{O}_{3}}$ dissociates and gives 2 ions, so $i=2$
Thus, osmotic pressure will be
$\pi =2\times 0.01\times R\times T$
$\pi =0.02RT$
(C) $500ml$ solution containing $0.3g$ urea
Urea does not dissociate, that is, it does not give any ions, so $i=1$
Also, the concentration is calculated as
$c=\frac{0.3\times 1000}{60\times 500}$
$c=0.01$
Thus, osmotic pressure will be
$\pi =1\times 0.01\times R\times T$
$\pi =0.01RT$
(D) $0.04N$$HCl$
As $HCl$ dissociates and gives 2 ions, so $i=2$
The relationship between normality and molarity is
$Molarity=\frac{Normality}{Valency}$
$Molarity=\frac{0.04}{1}$
$Molarity=0.04M$
Thus, osmotic pressure will be
$\pi =1\times 0.04\times R\times T$
$\pi =0.04RT$
Hence, $0.01M$ glucose and $500ml$ solution containing $0.3g$ urea are isotonic solutions.
Correct Options: (A) and (C).
Note: The cells can transport water and nutrients into and out of the cells due to the isotonic solution. This is necessary for blood cells to carry out their duty of supplying oxygen and other nutrients to other regions of the body.
Formula Used: The formula used to calculate the osmotic pressure of a solution is:
$\pi = icRT$; where $i$ = van’t hoff index
$c$ = Concentration of solute, $R$ = Gas constant, $T$ = Temperature (Kelvin), $\pi $= Osmotic pressure.
Complete Step by Step Answer:
For two solutions to be isotonic, they should have the same osmotic pressure. So, we will calculate the osmotic pressure of each of the solutions given in the options, and the two solutions that will have the same osmotic pressure will be our answer.
(A) $0.01M$ glucose
As glucose does not dissociates, that is, it does not give any ions, so $i=1$
Thus, osmotic pressure will be
$\pi =1\times 0.01\times R\times T$
$\pi =0.01RT$
(B) $0.01M$ $NaN{{O}_{3}}$
As $NaN{{O}_{3}}$ dissociates and gives 2 ions, so $i=2$
Thus, osmotic pressure will be
$\pi =2\times 0.01\times R\times T$
$\pi =0.02RT$
(C) $500ml$ solution containing $0.3g$ urea
Urea does not dissociate, that is, it does not give any ions, so $i=1$
Also, the concentration is calculated as
$c=\frac{0.3\times 1000}{60\times 500}$
$c=0.01$
Thus, osmotic pressure will be
$\pi =1\times 0.01\times R\times T$
$\pi =0.01RT$
(D) $0.04N$$HCl$
As $HCl$ dissociates and gives 2 ions, so $i=2$
The relationship between normality and molarity is
$Molarity=\frac{Normality}{Valency}$
$Molarity=\frac{0.04}{1}$
$Molarity=0.04M$
Thus, osmotic pressure will be
$\pi =1\times 0.04\times R\times T$
$\pi =0.04RT$
Hence, $0.01M$ glucose and $500ml$ solution containing $0.3g$ urea are isotonic solutions.
Correct Options: (A) and (C).
Note: The cells can transport water and nutrients into and out of the cells due to the isotonic solution. This is necessary for blood cells to carry out their duty of supplying oxygen and other nutrients to other regions of the body.
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