Experimental work (1) Answers

Answer to Experimental work Question 1using pV = nRTn = Mass/28.8 [1]Mass of 100 cm3 of air = (28.8 x 1.00 x 105 x 100 x 10−6) ÷ (8.31 x 293) = 0.118 g [1]orusing 1 mole of any gas occupies 22.7 dm3 at STP1 mole occupies 22.7 x (293 ÷ 273) = 24.4 dm3 at 293 K [1]Mass of 100 cm3 of air = (28.8 x 100) ÷ (24.4 x 1000) = 0.118 g [1](b) Mass of X = increase in mass + mass of 100 cm3 of air

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