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Important Question: ICSE 2010 Class 10th (Mole Concept and Stoichiometry A) Vapour Density and Molecular Weight. Num. 1. When heated, potassium permanganate decomposes according to the following equation : 2KMnO4 K 2 MnO4 + MnO2 + O2 Some potassium permanganate was heated in a test tube. After collecting one litre of oxygen at room temperature, it was found that the test-tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same condition of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen. Sol : - Mass of 1 lit. of oxygen = 1.32 g Mass of 1 lit. of hydrogen = 0.0825 g Vapour density of oxygen = mass of 1 lit. of oxygen / mass of 1 lit. of hydrogen = 1.32 /0.0825 = 16 g. Relative molecular mass of oxygen = 2 vapour density = 2 16 = 32. Q. 2. Name the term which defines the mass of a given volume of a gas compared to the mass of an equal volume of hydrogen. Ans.: - Vapour density. Num. 2. Find the relative molecular mass of a gas, 0.546 g of which occupies 360 cm3 at 87 C and 380 mm Hg pressure. [1 litre of hydrogen at s.t.p. weigh 0.09 g]. Sol : - P1 = 380 mm of Hg, V 1 = 360 cm3 = 360 ml, T1 = 87 + 273 = 360 K. P2 = 760 mm of Hg, V 2 = x ml, T2 = 273 K Using, P1 V1 /T1 = P2 V2 / T2 we get (380 360) / 360 = (760 x) / 273 => x = (273 380) / 760 = 136.5 l Mass of 136.5 ml = 0.546 g => mass of 1000 ml = (0.546/136.5) 1000 = 4 g Vapour density = mass of 1 lit. of the gas / mass of 1 lit. of hydrogen = 4/0.09 = 44.44 Molecular mass = 2 44.44 = 88.88 [Ans.] Percentage Composition. Num 1. Find the total percentage of oxygen in magnesium nitrate crystal : Mg(NO3)2.6H2 O [H = 1, N = 14, O = 16, Mg = 24]. Sol :- gram molecular mass of Mg(NO 3 )2.6H2O = 24 + (14 + 48) 2 + 6 18 = 256.
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