The average osmotic pressure of human blood is 7.8 bar at $37^\circ C$. What is the concentration of an aqueous NaCl solution that could be used in the blood steam?

$\pi = CRT:C = \frac{\pi }{{RT}} = \frac{{7.8}}{{0.0821 \times 310}} = 0.31\;{\rm{mol}}/{\rm{L}}$

The correct Option is "0.31 mol/L"

Equimolar solution in the same solvent have

Boiling point and freezing point depend on ${k_d}$ (molal elevation constant) and ${k_f}$ (molal depression constant) of the solvent. Thus, equimolar solution (of the non-electrolyte) will have same boiling point and also same freezing point. $\Delta {T_f} = {k_f} \times molality$ $\Delta {T_b} = {k_b} \times molality$ Note In this question nature of salute has not been mentioned. Hence, we have assumed that solute is non-electrolyte.

Different boiling and different freezing points

The correct Option is "Same boiling and same freezing points"

Same freezing point but different boiling point

Same boiling point but different freezing point

Blood cells retain their normal shapes in solutions which are :

Osmosis does not take place if two solutions are isotonic.

The correct Option is "Isotonic to blood"

At low concentrations, the statements that equimolal solutions under a given set of experimental conditions have equal osmotic pressure is true for

Equal osmatic pressure only applicable to non-electrolytes solution at low concentration

The correct Option is "Solutions of non-electrolytes only"

The van’t Hoff factor(i) for a dilute solution of ${{\rm{K}}_3}{\rm{Fe}}{\left( {{\rm{CN}}} \right)_6}$ is :

$\begin{array}{l} {{\rm{K}}_3}{\rm{Fe}}{\left( {{\rm{CN}}} \right)_6}\;\leftrightharpoons\;3{{\rm{K}}^ + } + {\rm{Fe}}\left( {{\rm{CN}}} \right)_6^{3 - }\\ \;\;\;\;\;\;1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0\;\;\;\;\;\;\;\;\;\;0\;\;\;\;\;\;\;\;\;\;\left( {Before\;dissociation} \right)\\ \;\;\;1 - \alpha \;\;\;\;\;\;\;\;\;\;\;\;\;3\alpha \;\;\;\;\;\;\;\;\;\alpha \;\;\;\;\;\;\;\;\;\left( {After\;dissociation} \right) \end{array}$ van’t Hoff factor $(i){\rm{ }} = {\rm{ }}1 - \alpha {\rm{ }} + 3\alpha {\rm{ }} + {\rm{ }}\alpha {\rm{ }} = {\rm{ }}1{\rm{ }} + {\rm{ }}3\alpha $ In very dilute state $\alpha = 1$; Thus, $i = 4$

Van’t Hoff factor of aq ${K_2}S{O_4}$ at infinite dillution has value equal to

${K_2}S{O_4} \to 2{K^ + } + SO_4^{2 - }$ It given 3 ions, hence , the van’t Hoff factor = 3.

A solution containing 500 g of a protein per litre is isotonic with a solution containing 3.42 g of sucrose per litre. The molecular mass of protein is :

For two non-electrolyte solutions to be isotonic; ${c_1} = {c_2},$ $\therefore \frac{{500}}{{m \times 1}} = \frac{{3.42}}{{342 \times 1}}$ $\therefore m = 50,000$

Osmotic pressure of 0.4% urea solution is 1.60 atm and that of 3.42% cane sugar is 2.46 atm. When the above two solutions are mixed, the osmotic pressure of the resulting solution is

Osmotic pressure of two solutions will be added. Hence, osmotic pressure of resulting solution =1.64+2.46 =4.10 atm.

The correct Option is "4.10 atm"

At $40^\circ C$ the vapour pressure in torr, of methyl alcohol-ethyl alcohol solutions is represented by the equation. $P = 119\;{X_A} + 135;$ where ${X_A}$ is mole-fraction of methyl alcohol, then the value of $\mathop {\lim }\limits_{{X_A} \to 1} \frac{{{P_A}}}{{{X_A}}}$ is :

$P = 119\;{X_A} + 135;\mathop {\lim }\limits_{{X_A} \to 1} \frac{{{P_A}}}{{{X_A}}} = 119 + 135 = 254\;{\rm{torr}}.$

The correct Option is "254 torr"

An 1% solution of KCl (I), NaCl (II), ${\rm{BaC}}{{\rm{l}}_2}$(III) and urea (IV) have their osmotic pressure at the same temperature in the ascending order (molar masses of NaCl, KCl, ${\rm{BaC}}{{\rm{l}}_2}$ and urea are respectively$58.5,{\rm{ }}74.5,{\rm{ }}208.4\;{\rm{ }}and\;{\rm{ }}60\;{\rm{ }}g\;{\rm{mo}}{{\rm{l}}^{ - 1}}$. Assume 100% ionisation of the electrolytes at this temperature

1% solution means 1 g solute is present in 100 mL of water. Osmotic pressure, $\pi = \frac{{iw \times RT}}{{M \times V}}$ ${\pi _{KCl}} = \frac{{2 \times 1 \times 1000 \times RT}}{{74.5 \times 100}}$ $ = 2 \times 0.134{\rm{ }}RT$ ${\pi _{NaCl}} = \frac{{2 \times 1 \times 1000 \times RT}}{{58.5 \times 100}}$ $ = 2 \times 0.171RT$ ${\pi _{BaC{l_2}}} = \frac{{3 \times 1 \times 1000 \times RT}}{{208.4 \times 100}}$ $ = 3 \times 0.048RT$ ${\pi _{urea}} = \frac{{1 \times 1 \times 1000 \times RT}}{{60 \times 100}}$ $ = 1 \times 0.167RT$ Since, temperature is same in all cases, the ascending order of osmotic pressure is $III < IV < I < II$