2 The questionIf 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?
3 Pull out the important parts 0.80 g of sulfur dioxide, SO2In 5.00 LAt 25◦CAt atm, P1X gIn 1 LAt 9.00 atm, P2
4 Start assembling the equation First, Calculate S1Solubility of a gas is measured in g/LSo we haveS1 = 0.80g5.00 L= 0.16 g / L
5 Solve for S2 Another way of looking at the equation S1 S2=P1 P2Cross multiply to get an equation that is easy to use.
6 Solve for S2 Another way of looking at the equation S1 S2=P1 P2Cross 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 P1We are solving for S2 so we’ll 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.S1P2P1(0.16g/L )(9.00 atm)10.00 atm= S2= g/L
8 It asked for how much will dissolve in 1 L. 0.144 gL= x g1 LHopefully 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?0.144 gL= x g0.75 LCross multiply to get (0.114 g) (0.75L) = (x g) (1 L)gL = x gLDivide both sides by L to get g will dissolve in 0.75 L