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3 months ago

Solutions Class 12th

When 12.2g of benzoic acid is dissolved in 100g of water, the freezing point of solution was found to be -0.93°C $[K_{f} (H_{2} O = 1.86 K k g m o l^{- 1}] .$ The number (n) of benzoic acid molecules associated (assuming 100% association) is…………,

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alok kumar singh

Contributor-Level 10

$Δ T_{f} = K_{f} * m * i$

0.93 = 1.86 * 1 * i

i = $\frac{1}{2}$

$i = 1 + (\frac{1}{n} - 1)$

$∴ n = 2$

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New answer posted

3 months 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________.

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Vishal Baghel

Contributor-Level 10

$Δ T_{b} = i K_{b} m$

$Δ T_{f} = i K_{f} m$

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

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

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3 months 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____________.

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Raj Pandey

Contributor-Level 9

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

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

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 phase, $y_{B} = \frac{P_{B}}{P_{A} + P_{B}} = \frac{70}{85}$

$\frac{70}{85} = \frac{x}{17}$

X = 14

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New answer posted

3 months ago

Solutions Class 12th

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alok kumar singh

Contributor-Level 10

$Δ T_{b} = i K_{b} m$

$Δ T_{f} = i K_{f} m$

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

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

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New answer posted

3 months 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]

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Raj Pandey

Contributor-Level 9

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

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

$Δ T_{f} = K_{f} * m$

3.9 * 0.386

$∴ x * 1 0^{- 1} = 15.05 * 1 0^{- 1}$

Ans = 15

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3 months 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 $K_{f} = 1.86 k k g m o l^{- 1}$ )

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Raj Pandey

Contributor-Level 9

Let we take $1 l$ of solution

Mass of solute = Volume * Density

= 0.5 $m l * 1.05 g m / m l$

= 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

3 months 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]

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Vishal Baghel

Contributor-Level 10

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

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

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

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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3 months ago

Solutions Class 12th

Determine the freezing point (nearest integer) of a 1 mole kg^-1 aqueous solution of a weak electrolyte that is 7.5% dissociated into two ions (in °C) [Given K f of water is 1.86 °C/m].

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New question posted

3 months ago

Solutions Class 12th

Calculate the strength in g/L of 10 volume solution of hydrogen peroxide at 273 K and 1 bar pressure. [Report your answer divided by 10]

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New answer posted

3 months ago

Solutions Class 12th

Consider a solution consisting of the following two buffer systems.

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Vishal Baghel

Contributor-Level 10

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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