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NEET Chemistry · 98 questions · Page 4 of 10 · Click an option or "Show Solution" to reveal answer

Q31
An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase?
A Addition of NaCl
B Addition of Na 2 SO 4
C Addition of 1.00 molal KI
D Addition of water
Correct Answer
Option D
Solution

Addition of solute decreases the vapour pressure as some sites of the surface are occupied by solute particles resulting in decreased surface area.

However, addition of solvent, i.e., dilution increases the surface area of the liquid surface, thus results in increased vapour pressure.

Hence, addition of water to the aqueous solution of (1 molal) KI results in increased vapour pressure.

Q32
A 0.0020 m aqueous solution of an ionic compound [Co(NH 3 ) 5 (NO 2 )]Cl freezes at −- 0.00732 o C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be ( K f = −-1.86 o C/m)
A 3
B 4
C 1
D 2
Correct Answer
Option D
Solution

The number of moles of ions produced by 1 mol of ionic compound = i Given, m = 0.0020 m

Δ\Delta

T f = 0 o C – 0.00732 o C = – 0.00732 o C K f = – 1.86 o C

Δ\Delta

T f = ik f m ⇒\Rightarrow i =

0.007321.86×0.0020{{0.00732} \over {1.86 \times 0.0020}}

= 2

Q33
0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If K f for water is 1.86 K kg mol −-1 , the lowering in freezing point of the solution is
A 0.56 K
B 1.12 K
C −- 0.56 K
D −- 1.12 K
Correct Answer
Option B
Solution

<table class=tg> <tbody><tr> <th class=tg-baqh></th> <th class=tg-nrix>HX</th> <th class=tg-nrix>⇌</th> <th class=tg-nrix>H<sup>+</sup></th> <th class=tg-nrix>+</th> <th class=tg-nrix>X<sup>-</sup></th> </tr> <tr> <td class=tg-baqh>Initially</td> <td class=tg-nrix>1</td> <td class=tg-nrix></td> <td class=tg-nrix>0</td> <td class=tg-nrix></td> <td class=tg-nrix>0</td> </tr> <tr> <td class=tg-baqh>At equilibrium</td> <td class=tg-nrix>1 - α\alpha</td> <td class=tg-nrix></td> <td class=tg-nrix>α\alpha</td> <td class=tg-nrix></td> <td class=tg-nrix>α\alpha</td> </tr> </tbody></table> <br><br>Total moles = 1 – α\alpha + α\alpha + α\alpha = 1 + α\alpha <br><br>∴\therefore i =

1+α1{{1 + \alpha } \over 1}

<br><br>Given, α\alpha = 20% = 0.2 <br><br>∴\therefore i = 1 + α\alpha = 1 + 0.2 = 1.2 <br><br>

Δ\Delta

T<sub>f</sub> = ik<sub>f</sub> m = 1.2 × 1.86 × 0.5 = 1.12 K

Q34
A solution of acetone in ethanol
A obeys Raoult's law
B shows a negative deviation from Raoult's law
C shows a positive deviation from Raoult's law
D behaves like a near ideal solution.
Correct Answer
Option C
Solution

A solution of acetone in ethanol shows positive deviation from Raoult's law.

It is because ethanol molecules are strongly hydrogen bonded.

When acetone is added, these molecules break the hydrogen bonds and ethanol becomes more volatile.

Therefore its vapour pressure is increased.

Q35
During osmosis, flow of water through a semipermeable membrane is
A from solution having lower concentration only
B from solution having higher concentration only
C from both sides of semipermeable membrane with equal flow rates
D from both sides of semipermeable membrane with unequal flow rates.
Correct Answer
Option D
Solution

The correct answer is Option D : from both sides of semipermeable membrane with unequal flow rates.

In osmosis, water (or any solvent) moves through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration.

This semipermeable membrane allows the passage of solvent molecules but not solute molecules.

The primary aim of this process is to equalize the solute concentrations on both sides of the membrane.

However, it is essential to note that while water molecules predominantly move from the area of lower concentration to the area of higher concentration, there is still some movement of water from the higher concentration to the lower concentration side as well.

This bidirectional movement occurs because water molecules are constantly in motion, but the net flow results in a higher volume of water moving to the side with higher solute concentration.

This process does not occur with equal flow rates because the driving force of osmosis is to reduce the concentration gradient between the two sides.

Therefore, the flow rate is predominantly higher from the lower solute concentration side to the higher solute concentration side until an equilibrium is reached, at which point the net flow of water equalizes, but does not necessarily stop.

This imbalance in flow rates demonstrates why Option D is the most accurate description of osmotic flow through a semipermeable membrane.

Q36
1.00 g of a non-electrolyte solute (molar mass 250 g mol −-1 ) was dissolved in 51.2 g of benzene. If the freezing point depression constant, K f of benzene is 5.12 K kg mol −-1 , the freezing point of benzene will be lowered by
A 0.2 K
B 0.4 K
C 0.3 K
D 0.5 K
Correct Answer
Option B
Solution

Molality of non-electrolyte solute =

1250×0.0512{1 \over {250 \times 0.0512}}

= 0.0781 m

Δ\Delta

T f = k f m = 5.12 × 0.0781 = 0.4 K

Q37
A solution containing 10 g per dm 3 of urea (molecular mass = 60 g mol −-1 ) is isotonic with a 5% solution of a nonvolatile solute is
A 200 g mol −-1
B 250 g mol −-1
C 300 g mol −-1
D 350 g mol −-1
Correct Answer
Option C
Solution

For isotonic solution, osmotic pressure of urea = osmotic pressure of nonvolatile solute

1060×1000{{10} \over {60 \times 1000}}

=

5m×100{5 \over {m \times 100}}

⇒\Rightarrow m = 300 g mol –1

Q38
A solution of urea (mol. mass 56 g mol −-1 ) boils at 100.18 o C at the atmospheric pressure. If K f and K b for water are 1.86 and 0.512 K kg mol −-1 respectively, the above solution will freeze at
A 0.654 o C
B −- 0.654 o C
C 6.54 o C
D −-6.54 o C
Correct Answer
Option B
Solution
Δ\Delta

T f = K f m ......(1)

Δ\Delta

T b = K b m ......(2) ⇒\Rightarrow

ΔTfΔTb=KfKb{{\Delta {T_f}} \over {\Delta {T_b}}} = {{{K_f}} \over {{K_b}}}

.....(3)

Δ\Delta

T b = T 2 – T 1 = 100.18 – 100 = 0.18 k f for water = 1.86 K kg mol –1 k b for water = 0.512 K kg mol –1

ΔTf0.18=1.860.512{{\Delta {T_f}} \over {0.18}} = {{1.86} \over {0.512}}

⇒\Rightarrow

ΔTf{\Delta {T_f}}

= 0.654

Δ\Delta

T b = T 2 – T 1 ⇒\Rightarrow 0.654 = 0 – T 2 ⇒\Rightarrow T 2 = −- 0.654 o C

Q39
A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at 20 o C are 440 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in the vapour phase would be
A 0.200
B 0.549
C 0.786
D 0.478
Correct Answer
Option D
Solution

As the ratio of pentane to hexane = 1 : 4 ∴\therefore Mole fraction of pentane = 1/5 Mole fraction of hexane = 4/5 Total vapour pressure =

(15×440+45×120)\left( {{1 \over 5} \times 440 + {4 \over 5} \times 120} \right)

= 184 mm of Hg ∴\therefore Vapour pressure of pentane in mixture = (Vapour pressure of mixture pentane ×\times Mole fraction of in vapour phase) ⇒\Rightarrow 88 = 184 × mole fraction of pentane in vapour phase ⇒\Rightarrow Mole fraction of pentane in vapour phase =

88184{{88} \over {184}}

= 0.478

Q40
The vapour pressure of two liquids P and Q are 80 and 60 torr, respectively. The total vapour pressure of solution obtained by mixing 3 mole of P and 2 mole of Q would be
A 72 torr
B 140 torr
C 68 torr
D 20 torr
Correct Answer
Option A
Solution

Hence total vapour pressure = [(Mole fraction of P) × (Vapour pressure of P)] + [(Mole fraction of Q) × Vapour pressure of Q)] =

(35×80+25×60)\left( {{3 \over 5} \times 80 + {2 \over 5} \times 60} \right)

= 48 + 24 = 72 torr

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