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Solution


Q.

Some CO2 gas was kept in a sealed container at a pressure of 1 atm and at 273 K. This entire amount of CO2 gas was later passed through an aqueous solution of Ca(OH)2. The excess unreacted Ca(OH)2 was later neutralised with 0.1 M of 40 mL HCl. If the volume of the sealed container of CO2 was x, then x is ______ cm3 (nearest integer).

[Given: The entire amount of CO2(g) reacted with exactly half the initial amount of Ca(OH)2 present in the aqueous solution.]                    [2025]

Ans.

(45)

Moles of HCl consumed=molarity of HCl×volume of HCl

                                                =0.1 M×40 mL=4 mmol

       Ca(OH)2+2HCl→CaCl2+2H2O

By stoichiometry, moles of Ca(OH)2 that reacted with HCl=Moles of HCl2 =4 mmol2=2 mmol

As out of total Ca(OH)2 present initially, half reacted with HCl and the other half reacted with CO2.

Moles of Ca(OH)2 reacted with CO2= Moles of Ca(OH)2 reacted with HCl=2 mmol

CO2+Ca(OH)2⟶CaCO3+H2O

By stoichiometry, moles of CO2=moles of Ca(OH)2 reacted with CO2=2 mmol=0.002 mol

At 1 atm and 273 K, 1 mol of any gas occupies 22400 cm3. So 0.002 mol of CO2 occupy 0.002×22400 cm3

                    =44.8 mL≈45 mL