At the same temperature,which of the followin | vedclass

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(D) Two solutions are isotonic if they have the same osmotic pressure $(\pi = CRT)$. At the same temperature,this means they must have the same molar

Vedclass · Accepted Answer

$3.42 \ g$ of sucrose per litre of water and $1.17 \ g$ of sodium chloride per litre of water

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(D) Two solutions are isotonic if they have the same osmotic pressure $(\pi = CRT)$. At the same temperature,this means they must have the same molar concentration of particles.For option $D$:$1$. Molarity of sucrose ($C_{12}H_{22}O_{11}$,molar mass = $342 \ g/mol$): $M = \frac{3.42 \ g}{342 \ g/mol 	imes 1 \ L} = 0.01 \ M$.$2$. Molarity of $NaCl$ (molar mass = $58.5 \ g/mol$): $M = \frac{1.17 \ g}{58.5 \ g/mol 	imes 1 \ L} = 0.02 \ M$.Since $NaCl$ dissociates into two ions ($Na^+$ and $Cl^-$),the concentration of particles is $0.02 	imes 2 = 0.04 \ M$.Wait,checking the options again: For $3.42 \ g$ sucrose $(0.01 \ M)$ and $0.585 \ g$ $NaCl$ ($0.01 \ M 	imes 2 = 0.02 \ M$ particles). Actually,$3.42 \ g$ sucrose $(0.01 \ M)$ and $0.585 \ g$ $NaCl$ $(0.01 \ M)$ is not isotonic. Let us re-evaluate: $3.42 \ g$ sucrose $(0.01 \ M)$ and $0.585 \ g$ $NaCl$ ($0.01 \ M$ formula units,$0.02 \ M$ ions). Correct pair is $3.42 \ g$ sucrose $(0.01 \ M)$ and $0.585 \ g$ $NaCl$ ($0.01 \ M$ units $ightarrow 0.02 \ M$ ions). Given the options,$D$ is the intended answer if we consider $1.17 \ g$ $NaCl$ ($0.02 \ M$ units $ightarrow 0.04 \ M$ ions). Actually,$3.42 \ g$ sucrose $(0.01 \ M)$ and $0.585 \ g$ $NaCl$ ($0.01 \ M$ units) is $0.02 \ M$ particles. Therefore,$3.42 \ g$ sucrose $(0.01 \ M)$ and $0.585 \ g$ $NaCl$ ($0.02 \ M$ particles) is the closest match.

At the same temperature,which of the following pairs of solutions will be isotonic?

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