Question

$CaC{l}_2.6H_{2}O\left(s\right)\rightleftharpoons CaC{l}_2\left(s\right)+6H_{2}O\left(g\right)K_p=6.4\times {10}^{-17}at{m}^6$
Excess solid $CaC{l}_2.6H_{2}O$ and $CaC{l}_2$ are taken in a container containing some water vapours at a pressure of $1.14torr$ at a particular temperature.

• A. $CaC{l}_2\left(s\right)$ acts as drying agent under given condition
• B. $CaC{l}_2\left(s\right)$ acts as hygroscopic substance given condition
• C. $CaC{l}_2.6H_{2}O\left(s\right)$ acts as effluoroscent substance
• D. Mass of $CaC{l}_2.6H_{2}O\left(s\right)$ increases due to some reaction

Answer 1

The correct option is C $CaC{l}_2.6H_{2}O\left(s\right)$ acts as effluoroscent substance

For the process:

$CaC{l}_2.6H_{2}O\left(s\right)\rightleftharpoons CaC{l}_2\left(s\right)+6H_{2}O\left(g\right)$

Equilibrium constant is given by:

$K_p=p_{H_{2}O}^6=6.4\times {10}^{-17}at{m}^6$

or $p_{H_{2}O}=0.002atm$

As $760torr=1atm$

$\therefore 1.14torr=1.14/760atm=0.0015atm$

When partial pressure of water reduces from $0.002atm$ to $0.0015atm$, the equilibrium will shift to forward direction so as to increase the partial pressure, according to Le-Chaterlier's principle.

Therefore $CaC{l}_2.6H_{2}O$ will lose water and act as effluorescent.

Option C is correct.