This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
Step 1: Identify the given values and the relevant law. Graham's Law of Diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. For a fixed distance or amount of gas, the time taken for diffusion is directly proportional to the square root of its molar mass. The formula is: (t_1)/(t_2) = sqrt((M_1)/(M_2)) Where: t_1 = time for gas 1 t_2 = time for gas 2 M_1 = molar mass of gas 1 M_2 = molar mass of gas 2 Given: Time for gas X (t_X) = 28.3 \, s Time for oxygen (t_O_2) = 20 \, s Atomic mass of Oxygen (O) = 16 Atomic mass of Sulphur (S) = 32 Step 2: Calculate the molar mass of oxygen (M_O_2). Oxygen gas is diatomic, so its formula is O_2. M_O_2 = 2 × atomic mass of O = 2 × 16 = 32 \, g/mol Step 3: Substitute the values into Graham's Law equation. Let gas 1 be gas X and gas 2 be oxygen. (t_X)/(t_O_2) = sqrt((M_X)/(M_O_2)) 28.3 \, s20 \, s = sqrt((M_X)/(32 \, g/mol)) Step 4: Solve for the molar mass of gas X (M_X). First, divide the times: 1.415 = sqrt((M_X)/(32)) Square both sides of the equation to eliminate the square root: (1.415)^2 = (M_X)/(32) 2.002225 = (M_X)/(32) Now, multiply by 32 to find M_X: M_X = 2.002225 × 32 M_X = 64.0712 \, g/mol Rounding to three significant figures (consistent with 28.3 \, s), the molecular mass of gas X is 64.1 \, g/mol. The molecular mass of gas X is 64.1 g/mol. That's 2 down. 3 left today — send the next one.