Solutions — Page 5 | NEET Chemistry

Q41

A solution contains non volatile solute of molecular mass M 2 . Which of the following can be used to calculate the molecular mass of solute in terms of osmotic pressure?

A

{M_2} = \left( {{{{m_2}} \over \pi }} \right)VRT

B

{M_2} = \left( {{{{m_2}} \over V}} \right){{RT} \over \pi }

C

{M_2} = \left( {{{{m_2}} \over V}} \right)\pi RT

D

{M_2} = \left( {{{{m_2}} \over V}} \right){\pi \over {RT}}

Correct Answer

Option B

Solution

For dilute solution, the van’t Hoff equation is

\pi = {n \over V}RT

$\Rightarrow$

\pi V = nRT

$\Rightarrow$

\pi V = {{{m_2}} \over M}RT

$\Rightarrow$

{M_2} = \left( {{{{m_2}} \over V}} \right){{RT} \over \pi }

Q42

A solution containing components A and B folloes Raoult's law

A A - B attraction force is greater than A - A and B - B

B A - B attraction force is less than A - A and B - B

C A - B attraction force remains same as A - A and B - B

D volume of solution is different from sum of volume of solute and solvent.

Correct Answer

Option C

Solution

These two components A and B follows the condition of Raoult’s law if the force of attraction between A and B is equal to the force of attraction between A and A or B and B.

Q43

Pure water can be obtained from sea water by

A centrifugation

B plasmolysis

C reverse osmosis

D sedimentation

Correct Answer

Option C

Solution

The osmotic pressure of sea water is 25 atm at 15°C.

When pressure greater than 26 atm is applied on sea water separated by a rigid semipermeable membrane.

Pure water is obtained.

This is also called desalination of sea water.

Q44

From the colligative properties of solution, which one is the best method for the determination of molecular weight of proteins and polymers?

A Osmotic pressure

B Lowering in vapour pressure

C Lowering in freezing point

D Elevation in boiling point

Correct Answer

Option A

Solution

Molecular masses of polymers are best determined by osmotic pressure method .

Firstly because other colligative properties give so low values that they cannot be measured accurately and secondly, osmotic pressure measurements can be made at room temperature and do not require heating which may change the nature of the polymer.

Q45

Which of the following aqueous solution will exhibit highest boiling point?

A

0.01 ~\mathrm{M} ~\mathrm{Na}_2 \mathrm{SO}_4

B

0.015 \mathrm{~M} ~\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6

C 0.01 M Urea

D

0.01 \mathrm{M} \mathrm{~KNO}_3

Correct Answer

Option A

Solution

The boiling point elevation of an aqueous solution can be determined using the formula: $\Delta \mathrm{T}_{\mathrm{b}} = i \mathrm{~K}_{\mathrm{b}} \times \mathrm{m}$ Where: $\Delta \mathrm{T}_{\mathrm{b}}$ is the change in boiling point. $i$ is the van’t Hoff factor (number of particles the solute breaks into). $\mathrm{K}_{\mathrm{b}}$ is the ebullioscopic constant of the solvent. $\mathrm{m}$ is the molality of the solution.

For simplicity, we'll assume molarity equals molality.

Here's the calculation for each solution: 0.01 M Na $_2$ SO $_4$ : $i = 3$ (Na $_2$ SO $_4$ dissociates into 2 Na $^+$ and 1 SO $_4^{2-}$ ) $i \times m = 3 \times 0.01 = 0.03$ 0.015 M C $_6$ HundefinedO $_6$ (glucose): $i = 1$ (glucose does not dissociate) $i \times m = 1 \times 0.015 = 0.015$ 0.01 M Urea : $i = 1$ (urea does not dissociate) $i \times m = 1 \times 0.01 = 0.01$ 0.01 M KNO $_3$ : $i = 2$ (KNO $_3$ dissociates into 1 K $^+$ and 1 NO $_3^-$ ) $i \times m = 2 \times 0.01 = 0.02$ The boiling point elevation is directly proportional to $i \times m$ .

Thus, the solution with the highest $i \times m$ value will have the highest boiling point.

Therefore, the 0.01 M Na $_2$ SO $_4$ solution, with an $i \times m$ value of 0.03, will have the highest boiling point.

Q46

If liquids A and B form an ideal solution

A the entropy of mixing is zero

B the free energy of mixing is zero

C the free energy as well as the entropy of mixing are each zero

D the enthalpy of mixing is zero

Correct Answer

Option D

Solution

When

A

and

B

from an ideal solution,

\Delta {H_{mix}} = 0

Q47

Consider a binary solution of two volatile liquid components 1 and

2 . x_1

and

y_1

are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of

\dfrac{1}{x_1}

vs

\dfrac{1}{y_1}

are given respectively as :

A

\dfrac{\mathrm{P}_1^0}{\mathrm{P}_2^0}, \dfrac{\mathrm{P}_1^0-\mathrm{P}_2^0}{\mathrm{P}_2^0}

B

\dfrac{\mathrm{P}_2^0}{\mathrm{P}_1^0}, \dfrac{\mathrm{P}_2^0-\mathrm{P}_1^0}{\mathrm{P}_2^0}

C

\dfrac{\mathrm{P}_2^0}{\mathrm{P}_1^0}, \dfrac{\mathrm{P}_1^0-\mathrm{P}_2^0}{\mathrm{P}_2^0}

D

\dfrac{\mathrm{P}_1^0}{\mathrm{P}_2^0}, \dfrac{\mathrm{P}_2^0-\mathrm{P}_1^0}{\mathrm{P}_2^0}

Correct Answer

Option D

Solution

For a binary solution of two volatile liquid components labeled 1 and 2, let $x_1$ and $y_1$ represent the mole fractions of component 1 in the liquid and vapor phases, respectively.

The linear relationship between the inverse of these mole fractions is plotted as $\dfrac{1}{x_1}$ versus $\dfrac{1}{y_1}$ .

To derive the slope and intercept of this linear plot, consider the following calculations: Using Raoult's Law for a Liquid Solution: For a liquid solution with volatile components 1 and 2: $\mathrm{P}_1 = \mathrm{P}_{\mathrm{T}} \cdot y_1 = \mathrm{P}_1^{\mathrm{o}} \cdot x_1$ Therefore: $\dfrac{\mathrm{P}_{\mathrm{T}}}{x_1} = \dfrac{\mathrm{P}_1^{\mathrm{o}}}{y_1}$ Rearranging the Equation: By substituting and rearranging, we have: $\dfrac{\mathrm{P}_2^{\mathrm{o}} + x_1(\mathrm{P}_1^{\mathrm{o}} - \mathrm{P}_2^{\mathrm{o}})}{x_1} = \dfrac{\mathrm{P}_1^{\mathrm{o}}}{y_1}$ Simplifying further: $\dfrac{\mathrm{P}_2^{\mathrm{o}}}{x_1} + (\mathrm{P}_1^{\mathrm{o}} - \mathrm{P}_2^{\mathrm{o}}) = \dfrac{\mathrm{P}_1^{\mathrm{o}}}{y_1}$ Expressing $\dfrac{1}{x_1}$ : Solving for $\dfrac{1}{x_1}$ , we obtain: $\dfrac{1}{x_1} = \left(\dfrac{\mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}\right)\left(\dfrac{1}{y_1}\right) + \left(\dfrac{\mathrm{P}_2^{\mathrm{o}} - \mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}\right)$ Determining the Slope and Intercept: The slope of the line is: $\text{Slope} = \dfrac{\mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}$ The intercept of the line is: $\text{Intercept} = \dfrac{\mathrm{P}_2^{\mathrm{o}} - \mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}$ In summary, for the plot of $\dfrac{1}{x_1}$ against $\dfrac{1}{y_1}$ , the slope is $\dfrac{\mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}$ and the intercept is $\dfrac{\mathrm{P}_2^{\mathrm{o}} - \mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}$ .

Q48

Benzene and toluene form nearly ideal solutions. At 20 oC, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at 20 oC for a solution containing 78 g of benzene and 46 g of toluene in torr is

A 50

B 25

C 53.5

D 37.5

Correct Answer

Option A

Solution

Given, Vapour pressure of benzene

= 75\,\,torr

Vapour pressure of toluene

= 22\,\,torr

mass of benzene in

=78g

hence moles of benzene

= {{78} \over {78}} = 1\,mole

(mol. wt of benzene

=78

) mass of toluene in solution

=46g

hence moles of toluene

= {{46} \over {92}} = 0.5\,\,mole

Total moles of benzene and toluene = 1.5 mol now partial pressure of benzene

= {P^ \circ }_b \times {X_b} = 75 \times {1 \over {1 + 0.5}}

= 75 \times {1 \over {1.5}} = 75 \times {2 \over 3} = 50

torr

Q49

A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution ?

A The solution formed is an ideal solution

B The solution is non-ideal, showing +ve deviation from Raoult’s law.

C The solution is non-ideal, showing –ve deviation from Raoult’s law.

D n-heptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s law.

Correct Answer

Option B

Solution

For this solution intermolecular interactions between

n

-heptane and ethanol aare weaker than

n

-heptane -

n

- heptane & ethanol-ethanol interactions hence the solution of

n

-heptane and ethanol is non-ideal and shows positive deviation from Raoult's law.

Q50

The degree of dissociation (

\alpha

) of a weak electrolyte, AxBy is related to van’t Hoff factor (i) by the expression :

A

\alpha = {{i - 1} \over {x + y + 1}}

B

\alpha = {{x + y - 1} \over {i - 1}}

C

\alpha = {{x + y + 1} \over {i - 1}}

D

\alpha = {{i - 1} \over {(x + y - 1)}}

Correct Answer

Option D

Solution

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{A_x}\,{B_y}\,\,\rightleftharpoons\,\,{}_x{A^{y + }} + {}_y{B^{x - }}

t = 0\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0

{t_{eq}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 - \alpha \,\,\,\,\,\,\,\,\,x\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\alpha

Total no. of moles

(i)

= 1 - \alpha + x\alpha + y\alpha

i - 1 = x\alpha + y\alpha - \alpha

= \alpha \left( {x + y - 1} \right)

$\therefore$

\,\,\,\,\,\,\,\alpha = {{i - 1} \over {\left( {x + y - 1} \right)}}