At the same temperature,which of the followin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Answer

(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?

Similar Questions

What is the relationship between the osmotic pressures of $6.00 \ g \ L^{-1}$ $CH_3COOH$ $(\pi_1)$ and $7.45 \ g \ L^{-1}$ $KCl$ $(\pi_2)$ solutions?

Find out the osmotic pressure of a $12 \% \ w/V$ $KCl$ solution at $27 \ ^oC$. (in $atm$)

. . . . . . solution is hypertonic with reference to fluid inside the blood cell.

$18 \ g$ glucose (Molar mass $= 180 \ g/mol$) is dissolved in $100 \ mL$ of water at $300 \ K$. If $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$,what is the osmotic pressure of the solution (in $atm$)?

When two solutions $X$ and $Y$ are separated by a semi-permeable membrane and the solvent flows from $Y$ to $X$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module