Q1: Pure oxygen is mixed with air to produce an enriched air containing 50 volume % of oxygen. The ratio of moles of air to oxygen used is

ANS:B - 0.58

To solve this problem, let's denote the volume of pure oxygen as 𝑉oxygenVoxygen​ and the volume of air as 𝑉airVair​. Since the enriched air contains 50 volume % of oxygen, it means that the volume of oxygen is half the total volume of the enriched air: 𝑉oxygen=0.5×(𝑉oxygen+𝑉air)Voxygen​=0.5×(Voxygen​+Vair​) Given that pure oxygen is mixed with air, we know that the volume of oxygen used is equal to the volume of pure oxygen used: 𝑉oxygen=𝑉pure oxygenVoxygen​=Vpure oxygen​ Let's denote the total volume of the mixture as 𝑉mixtureVmixture​, then we have: 𝑉mixture=𝑉oxygen+𝑉airVmixture​=Voxygen​+Vair​ From the first equation, we can express 𝑉oxygenVoxygen​ in terms of 𝑉airVair​: 𝑉oxygen=0.5×(𝑉oxygen+𝑉air)Voxygen​=0.5×(Voxygen​+Vair​) 𝑉oxygen=0.5×𝑉mixtureVoxygen​=0.5×Vmixture​ Since 𝑉oxygen=𝑉pure oxygenVoxygen​=Vpure oxygen​, we have: 𝑉pure oxygen=0.5×𝑉mixtureVpure oxygen​=0.5×Vmixture​ Now, let's consider the mole ratios: moles of oxygen=volume of pure oxygenmolar volume of oxygen at STPmoles of oxygen=molar volume of oxygen at STPvolume of pure oxygen​ moles of air=volume of airmolar volume of air at STPmoles of air=molar volume of air at STPvolume of air​ The ratio of moles of air to oxygen used is: moles of airmoles of oxygen=𝑉air𝑉molar volume of air0.5×𝑉mixture𝑉molar volume of oxygenmoles of oxygenmoles of air​=Vmolar volume of oxygen​0.5×Vmixture​​Vmolar volume of air​Vair​​​ At standard temperature and pressure (STP), the molar volume of oxygen is approximately 22.414 L/mol, and the molar volume of air (which is essentially the same as the molar volume of nitrogen, since air is mostly nitrogen) is also approximately 22.414 L/mol. Substituting these values: 𝑉air22.4140.5×𝑉mixture22.414=𝑉air0.5×𝑉mixture22.4140.5×Vmixture​​22.414Vair​​​=0.5×Vmixture​Vair​​ Now, using the fact that 𝑉mixture=𝑉oxygen+𝑉airVmixture​=Voxygen​+Vair​, we can rewrite 𝑉airVair​ as 𝑉mixture−𝑉oxygenVmixture​−Voxygen​: 𝑉mixture−𝑉oxygen0.5×𝑉mixture0.5×Vmixture​Vmixture​−Voxygen​​ =𝑉mixture0.5×𝑉mixture−𝑉oxygen0.5×𝑉mixture=0.5×Vmixture​Vmixture​​−0.5×Vmixture​Voxygen​​ =2−1=2−1 =1=1 So, the ratio of moles of air to oxygen used is 1. Therefore, none of the provided options match the correct ratio.