Question 13 Once you?ve determine the moles of gas that are generated and the mo

Question

Question 13 Once you?ve determine the moles of gas that are generated and the moles of gas that have not dissolved, you cna determine the moles of gas that have dissolved. Moles of gas generated = moles of gas dissolved + moles of gas not dissolved The solubility of CO2 is the moles of CO2 dissolved divided by the total volume in your buret (25.00 mL + V0 + V stopcock). A 0.16 g sample of CaCO3 is found to generate 21.0 mL of dry of CO2 gas at STP. If V01, and Vstopcock in your 25 mL buret are 3.81 mL and 1.71 mL respectively, what is the solubility of CO2? Your Answer: Answer units

Explanation / Answer

Answer – We are given, mass of CaCO3 = 0.16 g ,

Volume of CO2 = 21 mL at STP, so, P= 1.0 atm, T= 273 K

Vtop = 3.81 mL , Vstopcock = 1.71 mL

We are also given

Moles of gas generated = moles of gas dissolved + moles of gas not dissolved

Moles of CaCO3 = 0.16 g / 100.086 g.mol-1

                            = 0.0016 moles

CO2 gas mole generated

At STP

22.4 L = 1 moles

So, 0.021 L = ?

= 0.000938 moles of CO2

So moles of CO2 from the CaCO3 from theoretical 1:1 mole ratio

So, moles of CO2 dissolved = 0.0016 moles – 0.000938 moles

                                             = 0.000661 moles

Total volume = 25 +3.81 +1.71

                        = 30.52 mL

So solubility of CO2 = moles of CO2 / total volume

                                   = 0.000661 moles / 30.52 mL

                                    = 2.2*10-5 mol/mL

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