Question

Some CO\(_2\) gas was kept in a sealed container at a pressure of 1 atm and at 273 K. This entire amount of CO\(_2\) gas was later passed through an aqueous solution of Ca(OH)\(_2\). The excess unreacted Ca(OH)\(_2\) was later neutralized with 0.1 M of 40 mL HCl. If the volume of the sealed container of CO\(_2\) was \(x\), then \(x\) is ________________________ cm\(^3\) (nearest integer). [Given: The entire amount of CO\(_2\) reacted with exactly half the initial amount of Ca(OH)\(_2\) present in the aqueous solution.]

Hint:

When dealing with gas volume at standard conditions, use the molar volume of gas at STP, which is 22400 cm³ per mole.

Answer 1

Correct Answer: 45

Let the moles of CO\(_2\) be \(n\) and moles of Ca(OH)\(_2\) total initially be \(2n\). The excess Ca(OH)\(_2\) is \(n\). The gm equivalent of Ca(OH)\(_2\) is equal to the gm equivalent of HCl. Using the molarity and volume of HCl, we calculate: \[ n \times 2 = 0.1 \times \frac{40}{1000} \] \[ n = 2 \times 10^{-3} \] Now, the volume of CO\(_2\) is: \[ 2 \times 10^{-3} \times 22400 = 44.8 \, \text{cm}^3 \] Thus, the volume of CO\(_2\) is approximately 45 cm\(^3\).