For the equilibrium.SrCl2⋅6H2O(s)⇌SrCl2⋅2H2O(s)+4H2O(g)the equilibrium constant KP=16×10−12atm4 at 1∘C. If one litre of air saturated with water vapour at 1∘C is exposed to a large quantity of SrCl2⋅2H2O(s), what weight of water vapour will be absorbed? Saturated vapour pressure of water at 1∘C =7.6 torr.

Here, the pressure on the water molecule,PH2O4=16×10−12PH2O=2×10−3atmVapor pressure =7.6760=10−2atmw=(ΔP)VRT×M=10−2−2×10−3×10.0821×274×18=6.4mg

For the equilibrium.SrCl2⋅6H2O(s)⇌SrCl2⋅2H2O(s)+4H2O(g)the equilibrium constant KP=16×10−12atm4 at 1∘C. If one litre of air saturated with water vapour at 1∘C is exposed to a large quantity of SrCl2⋅2H2O(s), what weight of water vapour will be absorbed? Saturated vapour pressure of water at 1∘C =7.6 torr.

Here, the pressure on the water molecule,PH2O4=16×10−12PH2O=2×10−3atmVapor pressure =7.6760=10−2atmw=(ΔP)VRT×M=10−2−2×10−3×10.0821×274×18=6.4mg

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

For the equilibrium.
${SrCl}_{2} \cdot 6 H_{2} O (s) ⇌ {SrCl}_{2} \cdot 2 H_{2} O (s) + 4 H_{2} O (g)$
the equilibrium constant $K_{P} = 16 \times 10^{- 12} {atm}^{4}$ at $1^{\circ} C$ . If one litre of air saturated with water vapour at $1^{\circ} C$ is exposed to a large quantity of ${SrCl}_{2} \cdot 2 H_{2} O (s)$ , what weight of water vapour will be absorbed? Saturated vapour pressure of water at $1^{\circ} C$ =7.6 torr.

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

3.25 mg

8.5 mg

6.4 mg

2.3 mg

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Here, the pressure on the water molecule,

$P_{H_{2} O}^{4} = 16 \times 10^{- 12}$

$P_{H_{2} O} = 2 \times 10^{- 3} atm$

Vapor pressure $= \frac{7.6}{760} = 10^{- 2} atm$

$w = \frac{(ΔP) V}{RT} \times M = \frac{[10^{- 2} - 2 \times 10^{- 3}] \times 1}{0 .0821 \times 274} \times 18 = 6 .4 mg$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE