Question

Which of the following pairs of solutions are expected to be isotonic at the same temperature?
A.$0.2M$ urea and $0.2M$ $NaCl$
B.$0.1M$ urea and $0.2M$ $MgC{l_2}$
C.$0.10M$ $NaCl$ and $0.1M$ $N{a_2}S{O_4}$
D.$0.1M$ $Ca{\left( {N{O_3}} \right)_2}$ and $0.1M$ $N{a_2}S{O_4}$

Answer 1

Hint: We can predict which pair of solutions is isotonic by calculating the osmotic pressure of the solutions. We can calculate the osmotic pressure using the molarity of the solution, van’t Hoff factor, gas constant and pressure. We say the van't Hoff factor is the ratio between the original concentrations of particles obtained when the substance is soluble and the concentration of solution obtained from its mass.

Complete answer:
We have to know that for electrolytes, the van’t Hoff factor is the sum of the amount of ions that is found in one formula unit of the substances whereas for non-electrolytes, we can say that the van't factor is one.
We can say that expression to calculate the osmotic pressure could be given as,
\[\Pi = iMRT\]
We can say that,
i stands for van’t Hoff factor
$\Pi $ stands for osmotic pressure
R stands for gas constant
M stands for molarity
T stands for temperature (in Kelvins)
We have to know that, when two solutions have the same (or) equal osmotic pressure, they are said to be isotonic in nature.
Let us now determine the van’t Hoff factor and osmotic pressure for each pair of solutions.
The given pair is $0.2\,M$ urea and $0.2M$ $NaCl$ . In this pair, urea is non-electrolyte, so the van’t Hoff factor for urea is one whereas sodium chloride is an ionic compound, it has a total of two ions. So, its van’t Hoff factor is two. We can express the osmotic pressure as,
For $0.2\,M$ urea, the expression for osmotic pressure is $\Pi = \left( 1 \right)\left( {0.2} \right)RT$
For $0.2M$ $NaCl$ , the expression for osmotic pressure is $\Pi = \left( 2 \right)\left( {0.2} \right)RT$
The osmotic pressure could be different due to the difference in van’t Hoff factor. Option (A) is incorrect.
The given pair is $0.1\,M$ urea and $0.2M$ $MgC{l_2}$ . In this pair, urea is non-electrolyte, so the van’t Hoff factor for urea is one whereas magnesium chloride is an ionic compound, it has a total of three ions. So, its van’t Hoff factor is three. We can express the osmotic pressure as,
For $0.1\,M$ urea, the expression for osmotic pressure is $\Pi = \left( 1 \right)\left( {0.1} \right)RT$
For $0.2M$ $MgC{l_2}$ , the expression for osmotic pressure is $\Pi = \left( 3 \right)\left( {0.2} \right)RT$
The osmotic pressure could be different due to the difference in van’t Hoff factor. Option (B) is incorrect.
The given pair is $0.10M$$NaCl$ urea and $0.1M$ $N{a_2}S{O_4}$ . In this pair, sodium chloride is a strong electrolyte, so the van’t Hoff factor for sodium chloride is two whereas sodium sulfate is a strong electrolyte, it has a total of three ions. So, its van’t Hoff factor is three. We can express the osmotic pressure as,
For $0.10M$ $NaCl$ urea, the expression for osmotic pressure is $\Pi = \left( 2 \right)\left( {0.10} \right)RT$
For $0.1M$ $N{a_2}S{O_4}$ , the expression for osmotic pressure is $\Pi = \left( 3 \right)\left( {0.1} \right)RT$
The osmotic pressure could be different due to the difference in van’t Hoff factor. Option (C) is incorrect.
The given pair is $0.1M$ $Ca{\left( {N{O_3}} \right)_2}$ and $0.1M$ $N{a_2}S{O_4}$ . In this pair, calcium nitrate is a strong electrolyte, so the van’t Hoff factor for calcium nitrate is three whereas sodium sulfate is a strong electrolyte, it has a total of three ions. So, its van’t Hoff factor is three. We can express the osmotic pressure as,
For $0.1M$ $Ca{\left( {N{O_3}} \right)_2}$ the expression for osmotic pressure is $\Pi = \left( 3 \right)\left( {0.1} \right)RT$
For $0.1M$ $N{a_2}S{O_4}$ , the expression for osmotic pressure is $\Pi = \left( 3 \right)\left( {0.1} \right)RT$
The osmotic pressure could be the same due to the same value in the van't Hoff factor. Option (D) is correct.
We have identified the solution pair that is isotonic as $0.1M$ $Ca{\left( {N{O_3}} \right)_2}$ and $0.1M$ $N{a_2}S{O_4}$

Option (D) is correct.

Note:
We have to know that generally osmotic pressure is measured in terms of atmospheres. We have to know that the van't Hoff factor for organic compounds is generally one. Non-electrolytes as well as weak electrolytes are also one. Strong electrolytes have different van’t Hoff factors.