100 L (STP) of an O2-O3 mixture contains 20% O3 by volume. The mixture is passed through turpentine oil, which absorbs only O3 completely. Find the decrease in mass of the gas in grams, to the nearest integer (O = 16; 22.4 L/mol).

MolSynapse / Question bank / Mole Concept / Average molar mass of mixtures, vapour density

JEE Advanced levelNumerical answerDifficulty 3 of 3mc216

$100 L$ (STP) of an $O X_{2}$ - $O X_{3}$ mixture contains $20%$ $O X_{3}$ by volume. The mixture is passed through turpentine oil, which absorbs only $O X_{3}$ completely. Find the decrease in mass of the gas in grams, to the nearest integer (O = 16; 22.4 L/mol).

Numerical answer type, answer in g, tolerance 0.5.

Answer: 43 g

Volume of O3 = 20 L (volume % = mole %), so n(O3) = 20/22.4 = 0.8929 mol. Mass absorbed = 0.8929 x 48 = 42.86 g, i.e. 43 g to the nearest integer. (For reference the average M of the original mixture is 0.2 x 48 + 0.8 x 32 = 35.2.)

Practice Mole Concept More on Average molar mass of mixtures, vapour density

Previous questionNext question

100 L (STP) of an OX2​-OX3​ mixture contains 20% OX3​ by volume. The mixture is passed through turpentine oil, which absorbs only OX3​ completely. Find the decrease in mass of the gas in grams, to the nearest integer (O = 16; 22.4 L/mol).

Answer: 43 g

More questions on Average molar mass of mixtures, vapour density

$100 L$ (STP) of an $O X_{2}$ - $O X_{3}$ mixture contains $20%$ $O X_{3}$ by volume. The mixture is passed through turpentine oil, which absorbs only $O X_{3}$ completely. Find the decrease in mass of the gas in grams, to the nearest integer (O = 16; 22.4 L/mol).