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An ideal mixture of liquids A and B with 2 moles of A and 2 moles of B has a total vapour pressure of 1 atm at a certain temperature. Another mixture with 1 mole of A and 3 moles of B has a vapour pressure greater than 1 atm. But if 4 moles of C are added to the second mixture, the vapour pressure comes down to 1 atm. Vapour pressure of C, P_c° = 0.8 atm. Calculate the vapour pressures of pure A and pure B.

P_A° = 1.4 atm, P_B° = 0.7 atm

PA = 1.2 atm, P_B° = 0.6 atm

PA = 1.4 atm, P_B° = 0.6 atm

P_A° = 0.6 atm, P_B° = 1.4 atm

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Explanation

$\frac{P_{A}^{0}}{2} + \frac{P_{B}^{0}}{2}$ = 1 atm $\Rightarrow$ P_A° + P_B° = 2 atm

$\frac{P_{A}^{0}}{4} + \frac{{3P}_{B}^{0}}{4}$ > 1 atmrArrP_A° + 3P_B° > 4 atm

and $\frac{P_{A}^{0}}{8} + \frac{{3P}_{B}^{0}}{8}$ + $\frac{{4P}_{C}^{0}}{8}$ = 1 atmrArrP_A° + 3P_B° + 4P_C° = 8 atm

So P_A° + 3P_B° = (8 – 4 × 0.8 ) atm = 4.8 atm

Hence P_B° = 1.4 atm

P_B° = 0.6 atm

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A sample of air is saturated with benzene (vapor pressure = 100 mm Hg at 298 K) at 298K, 750mm Hgpressure. If it is isothermally compressed to one third of its initial volume, the final pressure of the system is

2250 torr

2150 torr

2050 torr

1950 torr

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Explanation

Pressure of air = 750 – 100 = 650 mm of Hg

on compressing P_f = 650 × 3 mm of Hg = 1950 mm of Hg

so P_T = (1950 + 100) = 2050 mm of Hg

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Available solutions are 1L of 0.1 M NaCl and 2L of 0.2 M CaCl₂. Using only these two solutions what maximum volume of a solution can be prepared having [Cl^–] = 0.34M exactly. Both electrolytes are strong

2.5 L

2.4 L

2.3 L

None of these

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Explanation

Let volumes taken to be ‘x’ & ‘y’ litres,

so $\frac{0 .1 x + 0 .4 y}{x + y}$ = 0.34&Vg = (x + y) (to be maximised)

so y = 4x so for maximum volume

y = 2L & x = $\frac{1}{2}$ L.

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Calculate the osmotic pressure of the solution prepared in the abovequestionT = 300 K, (R = 0.082 L atm mol^-1K^-1)

10.8 atm

12.8 atm

5.6 atm

None of these

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Explanation

π = CRT

C = (0.34 + $\frac{0 .1 \times 0 .5}{2 .5} + \frac{0 .2 \times 2}{2 .5})$ = 0.34 + 0.02 + 0.16 = 0.52

so π = 0.52 × 0.082 × 300 atm = 12.792 atm

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Consider equimolal aqueous solutions of NaHSO₄ and NaCl with AT_b and Ar_b as their respective boiling point elevations. The value of $\lim_{x \to 0} \frac{Δ T_{b}}{Δ T_{b}^{'}}$ will be :

1.5

3.5

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Explanation

As m → 0, NaHSO₄ will generate three particles while NaCl will generate only two particles.

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The vapor pressures of benzene, toluene and a xylene are 75 Torr, 22 Torrand 10 Torr at 20°C. Which of the following is not a possible value of the vapor pressure of an equimolar binary/ternary solution of these at 20°C ? Assume all form ideal solution with each other.

48 $\frac{1}{2}$

$35 \frac{2}{3}$

53 $\frac{1}{2}$

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Explanation

P_T = $\frac{1}{2}$ (75 + 22) = 48.5 torr

P_T = $\frac{1}{2}$ (75 + 10) = 42.5 torr

P_T = $\frac{1}{2}$ (22 + 10) = 16 torr

P_T = $\frac{1}{3}$ (75 + 22 + 10) = 35 $\frac{2}{3}$ torr

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15 g of methyl alcohol is dissolved in 35 g of water. The weight percentage of methyl alcohol in solution is

30%

50%

70%

75%

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Explanation

$(Weight percentage = \frac{Weight of solute}{Weight of solution} \times 100)$

Total weight of solution = (15 + 35) g = 50 g

Weight percentage of methyl alcohol $= \frac{Weight of methyl alcohol}{Weight of solution} \times 100 = \frac{15}{50} \times 100 = 30 %$

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Sea water contains 5.8 × 10^–3 g of dissolved oxygen per kilogram. The concentration of oxygen in parts per million is

5.8 ppm

58.5 ppm

0.58 ppm

0.05 ppm

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Explanation

Part per million $= \frac{Mass of solute}{Mass of solution} \times 10^{6} = \frac{5 .8 \times 10^{- 3} g}{10^{3} g} \times 10^{6}$ = 5.8 ppm

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A 500 gm toothpaste sample has 0.2 g fluoride concentration. The concentration of fluoride ions in terms of ppm level is [AIIMS 1994]

250 ppm

200 ppm

400 ppm

1000 ppm

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Explanation

ppm of F^– ions $= \frac{Mass of solute}{Mass of solution} \times 10^{6} = \frac{0 .2}{500} \times 10^{6}$ = 400 ppm

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Normality of a solution containing 9.8 g of H₂SO₄ in 250 cm³ of the solution is

[MP PMT 1995, 2003; CMC Vellore 1991; JIPMER 1991]

0.8 N

1 N

0.08 N

1.8 N

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Explanation

Eq. wt. of $H_{2} S O_{4} = \frac{Mol. mass of H_{2} S O_{4}}{Basicity of H_{2} S O_{4}} = \frac{98}{2} = 49$

∴ Number of g equivalent of H₂SO₄ $= \frac{Weight in g}{Eq. mass} = \frac{9 .8}{49} = 0 .2$

250 cm³ of solution contain H₂SO₄= 0.2 g equivalent

∴ 1000 cm³ of the solution contain H₂SO₄ $= \frac{0 .2}{250} \times 1000 g$ equivalent = 0.8 g equivalent

Hence normality of the solution = 0.8 N

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