Once you\'ve determine the moles of gas that are generated and the moles of gas

Question

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 CO_2 is the moles of CO_2 dissolved divided by the total volume In your buret (25.00 mL + V_top + V_StOpCOCk)- A 0.28 g sample of CaCO_3 is found to generate 23.3 ml of dry of CO_2 gas at STP. If V_stop and V_stopcock in your 25 mL buret are 3.96 ml and 2.03 ml respectively, what is the solubility of CO_2?

Explanation / Answer

Solution :-

CaCO3 ----- > CaO + CO2

Lets calculate the moles of CO2 that can be generated from 0.28 g CaCO3

(0.28 g CaCO3 * 1 mol / 100.0869 g )*(1mol CO2 / 1 mol CaCO3) = 0.002797 mol CO2

Now lets calculate the moles of dry CO2

(23.3 ml * 1 L / 1000 ml )*(1 mol / 22.4 L ) = 0.0010402 mol CO2

Now lets find the moles of CO2 that can be dissolced

Moles of CO2 dissolved = 0.002797 mol – 0.0010402 mol = 0.001756 mol

Total volume = 25 ml + 3.96 ml+ 2.03 ml = 30.99 ml = 0.03099 L

Solubility = moles dissolved / volume

                  = 0.001756 mol / 0.03099 L

                  = 0.0566 mol per L

So the solubility is 0.0566 mol per liter

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