The statement “If 0.003 mol of a gas are dis | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Henry’s law states that the mass of a gas dissolved in a given volume of liquid at a constant temperature is directly proportional to the pressure

Vedclass · Accepted Answer

Henry’s law

Vedclass · Answer

(D) Henry’s law states that the mass of a gas dissolved in a given volume of liquid at a constant temperature is directly proportional to the pressure of the gas in equilibrium with the liquid. Mathematically,$m \propto P$ or $S = k_H 	imes P$,where $S$ is the solubility and $P$ is the partial pressure. In this case,when pressure increases from $1 \ atm$ to $2 \ atm$ (doubles),the amount of dissolved gas increases from $0.003 \ mol$ to $0.006 \ mol$ (doubles). This direct proportionality confirms Henry’s law.

The statement “If $0.003 \ mol$ of a gas are dissolved in $900 \ g$ of water under a pressure of $1 \ atm$,$0.006 \ mol$ will be dissolved under a pressure of $2 \ atm$”,illustrates:

Similar Questions

Assertion $(A)$: For an endothermic dissolution process,an increase in temperature increases the solubility in a nearly saturated solution.
Reason $(R)$: In a saturated solution,dynamic equilibrium exists between the dissolved solute and the undissolved solute.

$H_2S$,a toxic gas with rotten egg like smell,is used for the qualitative analysis. If the solubility of $H_2S$ in water at $STP$ is $0.195 \, m$,calculate Henry's law constant.

What is the significance of Henry's Law constant $K_H$?

What is the solubility of oxygen gas in $100 \, cm^3$ of water at $293 \, K$ (in $, cm^3$)?

The solubility of a gas in a liquid increases with:

Vedclass Test Series

Exam Paper Generator

Online Exam Module