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JAMB UTME - Chemistry - 1990

Question 1 Report

The volume occupies by 1.58g of a gas at s.t.p is 500cm3. What is the relative molecular mass of the gas? (G.M.V at s.t.p = 22.40dm3)
Answer Details
The problem can be solved by using the ideal gas equation, which relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas: PV = nRT Where R is the universal gas constant. At STP (standard temperature and pressure), the temperature is 273.15 K and the pressure is 1 atm. Using the given volume of 500 cm3, we can convert it to dm3 by dividing by 1000: V = 500 cm3 = 0.5 dm3 Substituting these values into the ideal gas equation, we get: (1 atm) (0.5 dm3) = n (0.08206 L.atm/mol.K) (273.15 K) Simplifying this equation, we can solve for n: n = (1 atm) (0.5 dm3) / (0.08206 L.atm/mol.K) (273.15 K) n = 0.0200 mol Now we can use the formula for the density of a gas to find its relative molecular mass (M): density = mass / volume M = (mass of gas) / (number of moles of gas) We are given the mass of the gas (1.58 g) and the volume of the gas (0.5 dm3), so we can substitute these values into the density equation: density = (1.58 g) / (0.5 dm3) density = 3.16 g/dm3 Finally, we can solve for the relative molecular mass by rearranging the density equation: M = (density) (GMV) / (RT) Substituting the known values, we get: M = (3.16 g/dm3) (22.40 dm3/mol) / (0.08206 L.atm/mol.K) (273.15 K) M = 71.0 g/mol Therefore, the relative molecular mass of the gas is 71 g/mol, which corresponds to.