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Byju's Answer

Standard XII

Chemistry

Liquids in Liquids

CuSO4 · 5H2...

Question

$C u S O_{4}$ $\cdot$ $5 H_{2} O (s)$ $⇌$ $C u S O_{4} \cdot$ 3 $H_{2} O (s)$ + $2 H_{2} O (g)$ ; $K_{p}$ = 4 $\times 10^{- 4}$ $a t m^{2}$ if the vapour pressure of water is 38 torr then percentage of relative humidity is: ( Assume all data at constant temperature)

4

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10

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40

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n o n e o f t h e s e

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Solution

The correct option is C $40$
$K_{p} = P_{H_{2} O}^{2} = 4 \times 10^{- 4}$
$P_{H_{2} O} = 2 \times 10^{- 2} a t m \Rightarrow 0.02 a t m$
$P_{H_{2} O}^{1} = 38 a t m \Rightarrow \frac{38}{760} a t m \Rightarrow \frac{1}{20} a t m \Rightarrow 0.05 a t m$
$% Relative h u m i d i t y = \frac{P_{H_{2} O}}{P_{H_{2} O}^{1}} \Rightarrow \frac{0.02}{0.05} = \frac{2}{5} \times 100 \Rightarrow 40 %$
Option C is correct.

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Similar questions

{C u S O}_{4^{.}} {5 H}_{2} O (s) ⇌ {C u S O}_{4^{.}} {3 H}_{2} O (s) + 2 H_{2} O (g)

K_{p} = 4 \times 10^{- 4} {a t m}^{2}

and vapour pressure of water is

22.4

torr at

298

K. Then find out relative humidity.

Q. CuSO

_{4} .

H_{2} O

(s)

⇌

CuSO

_{4}

.3H

_{2} O

(s) + 2H

_{2} O

(g);

k_{p}

= 4

10^{4}

atm

^{2} .

If the vapour pressure of water is 38 torr then percentage of relative humidity is : (Assume all data at constant temperature)

Q. If

{C u S O}_{4} \cdot 5 H_{2} O (s) ⇌ {C u S O}_{4} \cdot {3 H}_{2} O (s) + {2 H}_{2} O (l) K_{p} = 1.086 \times 10^{- 4} {a t m}^{2} a t 25^{o} C

The efflorescent nature of

{C u S O}_{4} \cdot {5 H}_{2} O

can be noticed when vapour pressure of

H_{2} O

in atmosphere is:

10^{- 3}

mol of

C u S O_{4} \cdot 5 H_{2} O

is introduced in a

1.9 L

vessel maintained at a constant temperature of

27^{\circ} C

containing moist air at relative humidity of

12.5

%. What is the final molar composition of solid mixture?
For

C u S O_{4} .5 H_{2} O (s) ⇌ C u S O_{4} (s) + 5 H_{2} O (g), K_{p} (a t m) = 10^{- 10}

. Take vapor pressure of water at

27^{\circ} C

28

torrs.

If $C u S O_{4 (s)}$ . $5 H_{2} O_{(s)}$ $⇋$ $C u S o_{4}$ . $H_{2} O_{(s)}$ + $2 H_{2} O$ (g) $K_{p}$ = 1.0 x $10^{- 4}$ atm2 at $250^{\circ}$ . The efflorescent nature of CuSO4 . 5H2O can be noticed when vapour pressure of H2O in atmosphere is:

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CuSO4⋅5H2O(s)⇌ CuSO4⋅ 3H2O(s) + 2H2O(g); Kp = 4×10−4 atm2 if the vapour pressure of water is 38 torr then percentage of relative humidity is: ( Assume all data at constant temperature)

The correct option is C 40Kp=P2H2O=4×10−4PH2O=2×10−2atm⇒0.02atmP1H2O=38atm⇒38760atm⇒120atm⇒0.05atm%Relativehumidity=PH2OP1H2O⇒0.020.05=25×100⇒40%Option C is correct.

$C u S O_{4}$ $\cdot$ $5 H_{2} O (s)$ $⇌$ $C u S O_{4} \cdot$ 3 $H_{2} O (s)$ + $2 H_{2} O (g)$ ; $K_{p}$ = 4 $\times 10^{- 4}$ $a t m^{2}$ if the vapour pressure of water is 38 torr then percentage of relative humidity is: ( Assume all data at constant temperature)