Anhydrous calcium chloride is often used as a dessicant- In the presence of excess of CaCl2- the amount of the water taken up is governed by Kp-6-4 - 1085 for the following reaction at room temperature- CaCl2s-6H2Og- CaCl2-6H2Os- What is the equilibrium vapour pressure of water in a closed vessel that contains CaCl2s-

The correct option is D P_H_2O=5 × 10^ 15 atm CaCl_2(s)+6H_2O(g) → CaCl_2.6H_2O(s) For this reaction, K_p = 1/p_H_2O^6 = nbsp;6.4 × nbsp;10^8 ...

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Byju's Answer

Standard XII

Chemistry

Law of Mass Action

Anhydrous cal...

Question

Anhydrous calcium chloride is often used as a dessicant. In the presence of excess of $C a C l_{2}$ , the amount of the water taken up is governed by $K_{p} = 6.4 \times 10^{8} 5$ for the following reaction at room temperature, $C a C l_{2} (s) + 6 H_{2} O (g) ⇌ C a C l_{2} .6 H_{2} O (s) .$ What is the equilibrium vapour pressure of water in a closed vessel that contains $C a C l_{2} (s)$ ?

P_{H_{2} O} = 5 \times 10^{- 15} a t m

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P_{H_{2} O} = 6 \times 10^{- 15} a t m

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P_{H_{2} O} = 7 \times 10^{- 15} a t m

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P_{H_{2} O} = 9 \times 10^{- 15} a t m

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Solution

The correct option is D $P_{H_{2} O} = 5 \times 10^{- 15} a t m$
$C a C l_{2} (s) + 6 H_{2} O (g) \to C a C l_{2} .6 H_{2} O (s)$
For this reaction,
$K_{p} = 1 / p_{H_{2} O}^{6} = 6.4 \times 10^{85}; p_{H_{2} O} = 5 \times 10^{- 15} a t m .$

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Similar questions

Q. Anhydrous calcium chloride is often used as a desiccant. In the presence of excess of

C a C l_{2}

, the amount of the water taken up is governed by

K_{P} = 6.4 \times 10^{85}

for the following reaction at room temperature,

C a C l_{2} (s) + 6 H_{2} O (g) ⇌ C a C l_{2} . 6 H_{2} O (s) .

What is the equilibrium vapour pressure of water in a closed vessel that contains

C a C l_{2} (s) ?

C a C l_{2} .6 H_{2} O (s) ⇌ C a C l_{2} (s) + 6 H_{2} O (g) K_{p} = 6.4 \times 10^{- 17} a t m^{6}

Excess solid

C a C l_{2} .6 H_{2} O

and

C a C l_{2}

are taken in a container containing some water vapours at a pressure of

1.14 t o r r

at a particular temperature.

Q. For a given reaction

H_{2} O (l) ⇌ H_{2} O (g)

takes place in closed vessel at a constant temperature,

p_{H_{2} O}

is the vapour pressure of

H_{2} O

. The equilibrium constant,

K_{p}

is given by

On a certain humid day, the mole fraction of the water vapour in air at $25^{\circ} C$ is 0.0287. If total pressure of air is 0.977 bar, calculate the partial pressure of water vapour in air and relative humidity if the vapour pressure of water at $25^{\circ} C$ is 0.0313 bar. The partial pressure of water vapour ( $P_{H_{2} O}$ ) and relative humidity ( $R_{H}$ ) respectively are:

Q. Choose the correct expression of

K_{p}

for reaction,

4 N H_{3} (g) + 5 O_{2} (g) ⇌ 4 N O (g) + 6 H_{2} O (g)

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The correct option is D PH2O=5×10−15atmCaCl2(s)+6H2O(g)→CaCl2.6H2O(s)For this reaction,Kp=1/p6H2O=6.4×1085;pH2O=5×10−15atm.

The correct option is D $P_{H_{2} O} = 5 \times 10^{- 15} a t m$
$C a C l_{2} (s) + 6 H_{2} O (g) \to C a C l_{2} .6 H_{2} O (s)$
For this reaction,
$K_{p} = 1 / p_{H_{2} O}^{6} = 6.4 \times 10^{85}; p_{H_{2} O} = 5 \times 10^{- 15} a t m .$