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New answer posted•a year ago

Solutions Class 12th

The vapour pressure of two volatile liquid A and B at 25°C are 50 Torr and 100 Torr, respectively. If the liquid mixture contains 0.3 mole fraction of A, then the mole fraction of liquid B in the vapour phase is $\frac{x}{17} .$ The value of x is____________.

6 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

$P_{A}^{0} = 50 t o r r$

$P_{B}^{0} = 100 t o r r$

Mole fraction of A in liquid phase, x_A = 0.3

Mole fraction of B in liquid phase, x_B = 0.7

Now; $P_{A} = P_{A}^{0} x_{A}$

$P_{B} = P_{B}^{0} x_{B}$

= 100 × 0.7 = 70 torr

Mole fraction of B in vapour

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New answer posted•a year ago

Solutions Class 12th

Elevation in boiling point for 1.5 molal solution of glucose in water is 4 K. The depression in freezing point for 4.5 molal solution of glucose in water is 4K. The ratio of molal elevation constant to molal depression constant (K_b/K_f) is________.

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$Δ T_{b} = i K_{b} m$

$Δ T_{f} = i K_{f} m$

$∴ \frac{4}{4} = \frac{K_{b} \times 1.5}{K_{f} \times 4.5}$

$\frac{K_{b}}{K_{f}} = 3$

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New answer posted•a year ago

Solutions Class 12th

150g of acetic acid was contaminated with 10.2g ascorbic acid $(C_{6} H_{8} O_{6})$ to lower down its freezing point by (x × 10^-¹)°C. The value of x is _________. (Nearest integer)

(Given K_f = 3.9 K kg mol^-¹; molar mass of ascorbic acid = 176 g mol^-¹]

5 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

$m = \frac{w \times 1000}{m o l e c u l a r w t \times w_{s o l v e n t}} = \frac{10.2 \times 1000}{176 \times 150}$

$= \frac{10200}{176 \times 150} = 0.386$

$Δ T_{f} = K_{f} \times m$

3.9 × 0.386

$∴ x \times 1 0^{- 1} = 15.05 \times 1 0^{- 1}$

Ans = 15

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New answer posted•a year ago

Solutions Class 12th

The depression in freezing point observed for a formic acid solution of concentration 0.5 mL L^-1 is $0.0405 ° C .$ Density of formic acid is 1.05 g mL^-1. The Van’t Hoff factor of the formic acid solution is nearly: (Given for water $K_{f} = 1.86 k k g m o l^{- 1}$ )

8 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

Let we take $1 l$ of solution

Mass of solute = Volume × Density

= 0.5 $m l \times 1.05 g m / m l$

= 0.525 gram

Mass of solution = 1 kg. [considering very dilute solution]

Mass of solvent = 1000 – 0.525 = 999.475 gram

$\Rightarrow i = 1.9$

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New answer posted•a year ago

Solutions Class 12th

What is the apparent molecular mass (in gm/mol) of a weak acid, HA (molar mass = 30.0 gm/mol) in an aqueous solution, which shows elevation in boiling point by $0.0156 ° C ?$

[Given: pK_a (HA) = 2.0, K_b (H₂O) = 0.52 K kg mol^-1]

27 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$i = \frac{m_{t h e o r}}{m_{a p p}} = \frac{30}{m_{a p p}}$

$Δ T_{b} = i k_{b} . m] {H A}_{m - x}^{m} = {H^{+}}_{x}^{-} + {A^{-}}_{x}^{-}$

$0.0156 = (\frac{m \times x}{m}) \times 0.52 \times m$

m + x = 0.03

$\Rightarrow x = 0.01 \to i = 1.5 = \frac{30}{m_{a p p}} \Rightarrow m_{a p p p} = 20$

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New answer posted•a year ago

Solutions Class 12th

Consider a solution consisting of the following two buffer systems.

8 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

At pH = 6.4

As in case of H₂CO₃,

pH = pKa it will be only when

[weak acid] = [conjugate base].

In case of 2

H₂PO^-₄ |HPO^2-₄

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New answer posted•a year ago

Solutions Class 12th

Which of the following statement is true for an ideal solution?

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Solute and solvent both should obey Raoult's law

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New answer posted•a year ago

Solutions Class 12th

The density of a solution prepared by dissolving 1 gram molecule of sucrose in 1000 gram of water is 1.15 gm/ml. The molarity of the solution is ____.

4 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

Molarity $= \frac{moles of solute}{volume of solution in L} = \frac{1 * 1.15}{1342} * 1000 = 0.86 M$

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New answer posted•a year ago

Solutions Class 12th

In an estimation of sulphur by carius method, 0.25gm of an organic compound gave 0.35gm of BaSO₄. The mass percentage of sulphur in the compound is (Molecular weight of BaSO₄ = 233 g/mol )

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

∴ % of S in the compound
= (32/233) × (mass of BaSO? / mass of compound) × 100 = (32 × 0.35 × 100) / (233 × 0.25) = 19.227 ≈ 19.23

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New answer posted•a year ago

Solutions Class 12th

What volume in ml of 5 M H₂SO₄ should be added to 150ml of 1 M H₂SO₄ to obtain a solution of molarity = 2.5 M, if upon mixing volume of solution decreases by 10%?

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Let V ml is taken

$\frac{5 \times V + 1 \times 150}{(V + 150) + 0.9} = 2.5$

5V + 150 = 2.25 V + 337.5

$V = \frac{187.5}{2.75} = 68.18 m l$

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