Henry's constant (in kbar) for four gases α,β,γ and δ in water at 298 K is given below : α β γ δ K_H 50 2 2×10^(-5) 0.5 (density of water =10^3 k…

Question

Henry's constant (in kbar) for four gases α,β,γ and δ in water at 298 K is given below : α β γ δ K_H 50 2 2×10^(-5) 0.5 (density of water =10^3 k gm^(-3) at 298 K ) This table implies that :

MARKS Content Team · Accepted Answer

Correct answer: (C) The pressure of a 55.5 molal solution of δ is 250 bar. (1) From Henery's law P=k_HX Higher the value of k_H smaller will be solubility so γ is more soluble. (2) Though solubility of gas will decrease with increase in temperature but this conclusion can not be drawn from the given table. (3)For δ⇒P_(δ)=K_(Hδ).Xδ =0.5×10^3×55.55/(55.55+100018)=250bar. (4)For γ p_(γ)=KH_(γ )×x_(γ) =2×10^(-2)55.5/(55.5+100018)=10^(-2)bar

Henry's constant (in kbar) for four gases $α, β, γ$ and $δ$ in water at $298 K$ is given below :
$\begin{matrix} α & β & γ & δ \\ K_{H} & 50 & 2 & 2 \times 10^{- 5} & 0.5 \end{matrix}$
(density of water $= 10^{3} k {gm}^{- 3}$ at $298 K$ ) This table implies that :

Solution

Relevant Questions

Henry's constant (in kbar) for four gases α,β,γ and δ in water at 298 K is given below :αβγδKH5022×10-50.5(density of water =103 k gm-3 at 298 K ) This table implies that :

Solution

Relevant Questions

Henry's constant (in kbar) for four gases $α, β, γ$ and $δ$ in water at $298 K$ is given below :
$\begin{matrix} α & β & γ & δ \\ K_{H} & 50 & 2 & 2 \times 10^{- 5} & 0.5 \end{matrix}$
(density of water $= 10^{3} k {gm}^{- 3}$ at $298 K$ ) This table implies that :