A saturated solution forms when 0.0537 L of argon, at a pressure ...

Question

A saturated solution forms when 0.0537 L of argon, at a pressure 
of 1.0 atm and temperature of 25 °C, is dissolved in 1.0 L of 
water. Calculate the Henry's law constant for argon.

Accepted Answer

everyone's here, we have 0.352 liters of neon. And it's resolved in 1.1 liter of water at 0.96 atmosphere and 23 degrees Celsius. And were asked, what is Henry's law concept for neon when it produces a saturated solution. So we're gonna look at Henry's all, it's gonna be the solid bit of gas and hilarity. You can see the Henry's law constant similarity per atmosphere. That is the partial pressure of gas and we know that polarity, those moles of solute. What about leaders? Absolution? We need to first find moles. We're gonna use the ideal gas law to achieve this. It's gonna be P V N. R. T. We can rearrange this softer in. And he goes PV about about R. T. So in is what we're looking for was 0.96 atmosphere B 0. leaders and are 0.8206. Leader, sounds, atmosphere by moles times, kelvin and T. It's 23°C. We need to convert to Calvin. So we're gonna add 2073 15 And get 296.15 Kelvin. Now we're gonna pull this into the equation, we're going to get in Go 0.96 atmosphere. I'm 0.0352 Leaders By 0. atmosphere by Moles and Kelvin. That was 296.15 Kelvin. I'm gonna get 1.39 05 times 1093 moles. And I can rearrange Henry's law To solve Henry's law constant. Trying to get the malls of neon. What about leaders of the solution? Henry lost constant as the pressure pressure of neon K. H two Malls of Neon. Bye bye. Leader of the solution. The partial pressure of neon. We're gonna get 1.3905 cents omega three balls, About 1.1 L 0.96 atmosphere. So for Henry's law constant At 1.3 times 10 - three. Hilarity for atmosphere. Thanks for watching my video and I hope it was helpful.

A saturated solution forms when 0.0537 L of argon, at a pressure of 1.0 atm and temperature of 25 °C, is dissolved in 1.0 L of water. Calculate the Henry's law constant for argon.

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