What is Henry’s constant for neon dissolved in water given ${C_{Ne}} = 23.5mL\;{L^{ - 1}}$solution and STP volume $(22,414mL\;mol{e^{ - 1}}$gas) and pressure $(1atm)$?
Answer
624.9k+ views
Hint:As we know that Henry’s law states that the amount of the gas that dissolves in a given volume of liquid is directly proportional to the partial pressure of the gas in equilibrium with that solution, at a constant temperature. When this proportionality sign is removed the constant put in its place is the Henry’s constant which is a measure of the solubility of a gas.
Formula used: $moles = \dfrac{{Vol.\;of\;solute\;at\;STP}}{{22,414}}$ and $S = {K_H}P$.
Complete answer:
We know that according to Henry’s law which states that “the partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution”. When this proportionality sign is removed the constant put in its place is the Henry’s constant which is a measure of the solubility of a gas. This relation can be given as:
${p^\circ } = {K_H}\chi $ or in terms of solubility it can be written as $S = {K_H}P$ where ${K_H}$ is the Henry’s law constant.
Now we are given that $1$ litres of solution contains $23.5mL$ of Neon, so we can first calculate the moles of solution at STP. Using the formula:
$\Rightarrow moles = \dfrac{{Vol.\;of\;solute\;at\;STP}}{{22,414}}$
$\Rightarrow moles = \dfrac{{23.5}}{{22414}} = 0.001048\;moles$
So, we can say that the solubility in moles per litre or concentration of the solution is $0.001048M$.
Now using the above formula, we can calculate the Henry’s constant as:
$\Rightarrow S = {K_H}P$
$\Rightarrow 0.001048 = {K_H} \times 1$
Therefore the Henry’s constant is found to be ${K_H} = 0.001048\;L\;at{m^{ - 1}}$.
Note:Different gases have different Henry’s constant values at the same temperature suggesting that Henry’s constant is a function of nature of the gas. At a given pressure, higher the value of Henry’s constant, lower will be solubility and Henry’s constant increases with increase in temperature, so solubility will decrease with increase in temperature.
Formula used: $moles = \dfrac{{Vol.\;of\;solute\;at\;STP}}{{22,414}}$ and $S = {K_H}P$.
Complete answer:
We know that according to Henry’s law which states that “the partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution”. When this proportionality sign is removed the constant put in its place is the Henry’s constant which is a measure of the solubility of a gas. This relation can be given as:
${p^\circ } = {K_H}\chi $ or in terms of solubility it can be written as $S = {K_H}P$ where ${K_H}$ is the Henry’s law constant.
Now we are given that $1$ litres of solution contains $23.5mL$ of Neon, so we can first calculate the moles of solution at STP. Using the formula:
$\Rightarrow moles = \dfrac{{Vol.\;of\;solute\;at\;STP}}{{22,414}}$
$\Rightarrow moles = \dfrac{{23.5}}{{22414}} = 0.001048\;moles$
So, we can say that the solubility in moles per litre or concentration of the solution is $0.001048M$.
Now using the above formula, we can calculate the Henry’s constant as:
$\Rightarrow S = {K_H}P$
$\Rightarrow 0.001048 = {K_H} \times 1$
Therefore the Henry’s constant is found to be ${K_H} = 0.001048\;L\;at{m^{ - 1}}$.
Note:Different gases have different Henry’s constant values at the same temperature suggesting that Henry’s constant is a function of nature of the gas. At a given pressure, higher the value of Henry’s constant, lower will be solubility and Henry’s constant increases with increase in temperature, so solubility will decrease with increase in temperature.
Recently Updated Pages
Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

Father s age is three times the sum of the ages of-class-11-maths-CBSE

Give a comparative account of the classes of kingdom class 11 biology CBSE

The ceiling of a long hall is 25m high What is the class 11 physics CBSE

Name the Largest and the Smallest Cell in the Human Body ?

Draw a welllabelled diagram of a plant cell class 11 biology CBSE

Trending doubts
Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

