If 0.278g of argon dissolves in 1.5 l of water at 62 bar, what quantity of argon will dissolve at 78 bar

when P1/P2 = C1/C2

and C is the molarity which = moles/volume

so, P1/P2 = [(mass1/mw)/volume] / [(mass2/mw)/volume]

P1/P2 = (mass1/mw)/1.5L / (mass2/mw)/1.5L

so, Mw and 1.5 L will cancel out:

∴P1/P2 = mass1 / mass2

∴ mass 2 = mass1*(P2 / P1)

= 0.278g * (78 bar / 62 bar)

= 0.35 g

∴ the quantity of argon that will dissolve at 78 bar = 0.35 g

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nikitarahangdaleVT nikitarahangdaleVT · Answer 2 · 2022-03-19T07:39:59+01:00

0.35 grams of Argon gas will dissolve at 78 bar pressure in 1.5 liters of water.

What is ideal gas equation?

Ideal gas equation for any gas at standard condition will written as:

PV = nRT, where

P = pressure

V = volume

n = no. of moles

R = universal gas constant

T = temperature

For this equation volume and temperature is constant, so the required equation will be:

P₁/P₂ = n₁/n₂, where

P₁ = given pressure = 62 bar

n₁ = given mass of argon = 0.278g

P₂ = resultant pressure of argon = 78 bar

n₂ = to find?

On putting all these value on the above equation, we get

n₂ = 0.278g × (78 bar / 62 bar) = 0.35 g

Hence, required mass of argon is 0.35g.

To know more about ideal gas equation, visit the below link:

https://brainly.com/question/12873752