1. In an experiment 6.67 g of AlCl3 was produced and 0.54 g of Al remained unreacted.How many moles of Al were taken originally?
2.One mole of Potassium is themally decomposed and excess of aluminium is burnt in the gaseous product.How many moles of aluminium oxide are formed?z
3.A sample of H2SO4 is 90% by weight.What is the volume of the acid that can be used to make 1 litre of 0.2 M H2SO4
4.The density of 3M solution of Na2S2O3 is 1.2 g/ml.Find the molalities of Na+ and S2O3^2- ions.
5.If x% of a solute of molecular mass is present in a solution having density of d gm/cm^3 calculate molarity and molality if
a. x% is by mass b.x% is by volume
Asked by mehrarora26 | 22nd May, 2015, 09:43: PM
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Solution for your first query,
Al + 3/2 Cl2 ----> AlCl3
From the above reaction, 1mole Al produces 1 mole AlCl3
Finally 0.54 g Al is remaining that means 0.54 / 27 = 0.02 moles Al is remaining
6.67 g of AlCl3 is produced that means 6.67 / 133.5 = 0.05 moles of AlCl3 is produced.
From the reaction, 0.05 moles of AlCl3 can get produced from 0.05 moles of Al.
Hence, 0.05 moles used Al + 0.02 moles remaining Al = 0.07 moles of Al were present originally.
Answered by Prachi Sawant | 23rd May, 2015, 08:53: PM
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