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Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040.(assume the density of water to be one).


Mole fraction of ethanol is given by

Our aim is to find the number of moles of ethanol in 1L of the solution which is nearly = 1L of water.
Number of moles in 1L of water
                     
Substituting the values of  in eq. (1) we have
   

Hence molarity of the solution = 2.31 M
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Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
straight N subscript 2 left parenthesis straight g right parenthesis plus straight H subscript 2 left parenthesis straight g right parenthesis space rightwards arrow space space 2 NH subscript 3 left parenthesis straight g right parenthesis
(i) Calculate the mass of ammonia produced if 2·00 × 103 g dinitrogen reacts with 1·00×103 g of dihydrogen.
(ii) Will any of the two reactants remain un reacted?
(iii) If yes, which one and what would be its mass?

(i) 1 mol of N2 = 28g ; 3 mol of H2 = 6g;  28g of N2 react with 6g of H2
    
2000 of N2 would react with H2
                               
Thus, N2 in the limiting reagent while H2 is the excess reagent
2 mol of N2 i.e. 28g of N2 produce NH3
                       = 2 mol = 2 x 17 = 34g
 2000g of N2 would produce
             NH3

(ii) H2 will remain unreacted.

(iii) Mass of H2 that remains unreacted = 1000 - 428.6 = 571.4 g.
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In a reaction
straight A plus straight B subscript 2 space rightwards arrow space AB subscript 2
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B

(iii) 100 atoms of A + 100 molecules of B

(iv) 5 mol A + 2-5 mol B

(v) 2·5 mol A + 5 mol B


According to the given reaction:

(i) 1 atom of A will react with molecules of B2 = 1
300 atoms of A will react with molecules of B2 = 300
But molecules of B2 actually available = 200
∴  B2 is the limiting reactant.

(ii) 1 mol of A reacts with 1 mol of B
∴ 2 mol of A will react with 2 mol of B.
Hence A is the limiting reactant.

(iii) 100 atoms of A will react with 100 molecules of B. Hence there are no limiting reactants.

(iv) 2·5 mol of B will react with 2·5 mol of A. Hence B is the limiting reagent.

(v) 2·5 mol of A will react with 2·5 mol of B.

Hence A is the limiting reagent.

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What will be the mass of one straight C presuperscript 12 atom in g?


1 mol of  atoms  =  = 12 g
 

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How are 0.50 mol Na2CO3 and 0·50 M Na2CO3 different?


Molar mass of Na2CO3 = 2 x 23 + 12 + 3 x 16 = 106 g
0.50 mol Na2CO3 means 0.50 x 106g = 53g
0.50M Na2CO3 means 0.50 mol i.e. 0.50 x 106g, i.e. 53 g of Na2CO3 are present in 1 litre of the solution.

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