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

Explanation

Here's how to find the weight of liquefied oxygen:Convert volume to moles:○ We need to convert the volume of oxygen from milliliters (mL) to liters (L).○ 1 mL = 10^-3 L, so 112 mL = 112 x 10^-3 L = 0.112 L.Apply the ideal gas law (assuming ideal behavior for oxygen at STP):○ At STP (Standard Temperature and Pressure, 0°C and 1 atm), 1 mole of an ideal gas occupies 22.4 liters.○ Therefore, the number of moles of oxygen is:■ Moles of oxygen = Volume / Standard molar volume = 0.112 L / 22.4 L/mol ≈ 0.005 moles.Calculate the mass of oxygen using its molar mass:○ Molar mass of oxygen (O2) = 32 g/mol (from periodic table).○ Mass of oxygen = Moles of oxygen * Molar mass = 0.005 moles * 32 g/mol = 0.16 g.Therefore, the weight of 112 ml of oxygen at STP on liquefaction would be 0.16 g