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

CBSE Class 11-science Answered

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 | 22 May, 2015, 09:43: PM

Expert Answer

Dear mehrarora26@gmail.com

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.

In case of multiple questions within a query, please post each question individually and let us know where you are getting stuck so that we would be able to explain things better.

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.

Regards

Topperlearning Team.

Answered by Prachi Sawant | 23 May, 2015, 08:53: PM