Wt. of 112 ml of oxygen at STP on liquefaction would be

Explanation

Here's how to solve the problem:Convert volume to moles:We are given the volume of oxygen at STP: 112 mL.We need to convert this volume to moles because the mass depends on the number of moles, not just the volume.At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies a volume of 22.4 liters (L).First, convert the volume from milliliters (mL) to liters (L): 112 mL * (1 L / 1000 mL) = 0.112 L.Now, divide the volume by the volume per mole at STP: 0.112 L / 22.4 L/mol ≈ 0.005 mol.Calculate mass using molar mass:The molar mass of oxygen (O₂) is approximately 32 g/mol, meaning 32 grams of O₂ contain 1 mole of O₂ molecules.Knowing the number of moles (0.005 mol) and the molar mass, we can calculate the mass of O₂:Mass = Number of moles * Molar massMass = 0.005 mol * 32 g/mol ≈ 0.16 gTherefore, the weight of 112 mL of oxygen at STP on liquefaction would be 0.16 grams.