STD 11 - 4. Chemical bonding and molecular structure — NEET STD 11 Science — Question

$n - Pr = n -$ propyl
Product' $(A)$ is

$A.6{H_2}{O_{\left( s \right)}} \rightleftharpoons A.2{H_2}{O_{(s)}} + \mathop {4{H_2}{O_{(g)}}};{{K_P} = 1.6 \times {{10}^{ - 11}}} $

$B.12{H_2}{O_{\left( s \right)}} \rightleftharpoons B.7{H_2}{O_{(s)}} + \mathop {5{H_2}{O_{(g)}}};{{K_P} = 2.43 \times {{10}^{ - 13}}} $

$B.10{H_2}{O_{\left( s \right)}} \rightleftharpoons {C_{(s)}} + \mathop {10{H_2}{O_{(g)}}};{{K_P} = {{10}^{ - 30}}} $

Aqueous tension of $H_2O$ at $0\,^oC$ is given as $0.76\  torr$

$(I)$ The most effective drying agent will be $C_{(s)}$. Out of $C_{(s)}$ , $B. 7H_2O_{(s)}$ & $A. 2H_2O_{(s)}$

$(II)$ At $0\,^oC$, $A.6H_2O_{(s)}$ & $B.12H_2O_{(s)}$ will

$(III)$ If $R.H$ is less than $100\%$ in a chamber at $0\,^oC$, then none of the substance can act as deliquescent