At the same temperature, the following solution will be isotonic
A. \[3.24\text{ }gm~\] of sucrose per litre of water and \[0.18\text{ }gm~\] glucose per litre of water
B. \[3.42\text{ }gm~\] of sucrose per litre and \[0.18\text{ }gm~\] glucose in 0.1 litre of water
C. \[3.24\text{ }gm~\]of sucrose per litre of water and \[0.585\text{ }gm~\] of sodium chloride per litre of water
D. \[3.24\text{ }gm~\]of sucrose per litre of water and \[1.17\text{ }gm~\] of sodium chloride per litre of water

Hint: Isotonic solutions are those solutions when the osmotic pressure of the solutions is the same or can say the concentration of solute molecules in the solution is the same with respect to each other. In this question, we need to find which pair of solutions are isotonic. Osmotic pressure is represented as π and it is equal to iCRT. Where i is van’t hoff factor, C concentration of solute molecules in the solution, T is particular temperature and R is gas constant.

Complete Step by Step Answer:
In option A, a solution is a sucrose (\[3.24\text{ }gm~\]) and another solution is glucose (\[0.18\text{ }gm~\]). Let the osmotic pressure of sucrose solution is π1 and the osmotic pressure of glucose solution beπ2 such as
Let
\[\pi 1\text{ }=\pi 2\]
\[iCRT\text{ }=\text{ }iCRT\]
(as sucrose and glucose are non-electrolytes so \[i\text{ }=\text{ }1\])
\[CRT\text{ }=\text{ }CRT\]
(Concentration is defined as the number of moles of solute per litre of volume and in this option volume of solution is 1 litre. The moles is the ratio of mass in gram to the molar mass of solute)
\[3.24/342\text{ }\times \text{ }RT\text{ }/\text{ }1L\]which is not equal to
\[0.18/180\text{ }\times \text{ }RT/1L\]

In option B, one solution is sucrose (\[3.42\text{ }gm~\]) and the other is glucose (\[0.18\text{ }gm~\]).
Let
\[\pi 1\text{ }\left( osmotic\text{ }pressure\text{ }of\text{ }sucrose \right)\text{ }=\pi 2\text{ }\left( osmotic\text{ }pressure\text{ }of\text{ }glucose \right)\]
\[iCRT\text{ }=\text{ }iCRT\]
(as sucrose and glucose are non-electrolytes so \[i\text{ }=\text{ }1\])
\[CRT\text{ }=\text{ }CRT\]
(Given mass of sucrose is \[3.42\text{ }gm~\]and molecular mass of sucrose is 342. Volume of sucrose solution is 1 L. Given mass of glucose is \[0.18\text{ }gm~\] and molecular mass is 180. Volume of glucose solution is 0.1 L.)
\[3.42/342\text{ }\times \text{ }RT\text{ }/\text{ }1L\text{ }=\text{ }0.18/180\text{ }\times \text{ }RT/0.1L\]
\[0.01\text{ }=\text{ }0.01\]

In option C, one solution is sucrose (\[3.24\text{ }gm~\]) and the other is sodium chloride (\[0.585\text{ }gm~\]).
Let
\[\pi 1\text{ }\left( osmotic\text{ }pressure\text{ }of\text{ }sucrose \right)\text{ }=\pi 2\text{ }\left( osmotic\text{ }pressure\text{ }of\text{ }sodium\text{ }chloride \right)\]
\[iCRT\text{ }=\text{ }iCRT\]
(as sucrose is not electrolyte so \[i\text{ }=\text{ }1\] and sodium chloride is an electrolyte so i = 2)
\[CRT\text{ }=\text{ }2CRT\]
(Given mass of sucrose is \[3.24\text{ }gm~\]and molecular mass of sucrose is 342. Volume of sucrose solution is 1 L. Given mass of sodium chloride is \[0.585\text{ }gm~\] and molecular mass is \[58.5\]. Volume of sodium chloride solution is 0.1 L)
\[3.42/342\text{ }\times \text{ }RT\text{ }/\text{ }1L\text{ }is\text{ }not\text{ }equal\text{ }to\text{ }2\times \text{ }0.585/58.5\text{ }\times \text{ }RT/1L\]

In option D, one solution is sucrose (\[3.24\text{ }gm~\]) and the other is sodium chloride (\[1.17\text{ }gm~\]).
Let
\[\pi 1\text{ }\left( osmotic\text{ }pressure\text{ }of\text{ }sucrose \right)\text{ }=\pi 2\text{ }\left( osmotic\text{ }pressure\text{ }of\text{ }sodium\text{ }chloride \right)\]
\[iCRT\text{ }=\text{ }iCRT\]
(as sucrose is no electrolyte so \[i\text{ }=\text{ }1\]and sodium chloride is an electrolyte so \[i\text{ }=\text{ }2\])
\[CRT\text{ }=\text{ }2CRT\]
(Given mass of sucrose is \[3.24\text{ }gm~\]and molecular mass of sucrose is 342. Volume of sucrose solution is 1 L. Given mass of sodium chloride is \[1.17\text{ }gm~\] and molecular mass is \[58.5\]. Volume of sodium chloride solution is 0.1 L)
\[3.42/342\text{ }\times \text{ }RT\text{ }/\text{ }1L\text{ }is\text{ }not\text{ }equal\text{ }to\text{ }2\text{ }\times \text{ }1.17/58.5\text{ }\times \text{ }RT/1L\]
Thus, the correct option is B.

Note: In this question, we just find the concentration of solute molecules whose amount is given in grams. Concentration of solute is equal to the number of moles of solute (n) divided by volume of solution in litre (L). In this question volume of solution which is water is given in litre and we need to find the number of moles of solute. Number of moles is equal to the given mass divided by molecular mass. Molecular mass of sucrose 342, molecular mass of glucose is 180, molecular mass of sodium chloride is 58.5.