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S1S2=P1P2S1S2=P1P2 Henrys Law. If 0.80 g of sulfur dioxide at 10.00 atm pressure (P1) dissolves in 5.00 L of water at 25.0°C, how much of it will dissolve.

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Presentation on theme: "S1S2=P1P2S1S2=P1P2 Henrys Law. If 0.80 g of sulfur dioxide at 10.00 atm pressure (P1) dissolves in 5.00 L of water at 25.0°C, how much of it will dissolve."— Presentation transcript:

1 S1S2=P1P2S1S2=P1P2 Henrys Law

2 If 0.80 g of sulfur dioxide at atm pressure (P1) dissolves in 5.00 L of water at 25.0°C, how much of it will dissolve in 1 L of water at 9.00 atm pressure (P2) and the same temperature? The question

3 0.80 g of sulfur dioxide, SO2 In 5.00 L At 25C At atm, P1 X g In 1 L At 25C At 9.00 atm, P2 Pull out the important parts

4 First, Calculate S1 Solubility of a gas is measured in g/L So we have S1 = 0.80g 5.00 L = 0.16 g / L Start assembling the equation

5 Solve for S2 Another way of looking at the equation S1S2=P1P2S1S2=P1P2 Cross multiply to get an equation that is easy to use.

6 Solve for S2 Another way of looking at the equation S1S2=P1P2S1S2=P1P2 Cross multiply to get an equation that is easy to use. S1 P2 = S2 P1

7 Solve for S2 Another way of looking at the equation S1 P2 = S2 P1 We are solving for S2 so well divide both sides by P1. Yes, we could have just multiplied both sides by P2. I am trying to think of ways to make this equation easy to remember- is it easier as S1P2 = S2P1 or as S1 over P1 = S2 over P2? Use what makes your brain happy. S1P2 P1 = S2 (0.16g/L )(9.00 atm) atm = g/L

8 It asked for how much will dissolve in 1 L g L = x g 1 L Hopefully the answer is obvious that g will dissolve in 1 L. Just for fun, what if they asked how much would dissolve in 0.75L? g L = x g 0.75 L Cross multiply to get(0.114 g) (0.75L) = (x g) (1 L) gL = x gL Divide both sides by L to get g will dissolve in 0.75 L


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