Question

Q.63. An oleum sample is labelled as 118%, Calculate
Mass of H, SO, in 100 gm oleum sample.
Maximum mass of H,SO, that can be obtained if 30 gm sample is taken.
Composition of mixture (mass of components) if 40 gm water is added to 30 gm given
oleum sample.​

Answer 1

Answer:

Oleum si having the formulae (H2S2O7) and this is the mixture of H2SO4 and SO3.  

Then we can write the equation as H2S2O7 + H2O -> 2 H2SO4.

The molar mass of oleum is 178g, of water is  18g and form sulphuric acid will be 196g .

As, the 118% of oleum gives that  H2SO4 is 100g and SO3 is 18 g.  

We know from the equation that 1 mole of oleum will give 2 mole of sulfuric acid then, 118 g of oleum, the amount of H2SO4 produced will be 129.93 g.

Now, form the question we know that if 118 g of oleum, has 100 g of H2SO4 , then  according to the question 30 g of oleum will give (30 x 196) / 118 = 49.83 g of H2SO4.

As here 30g of oleum mixes with 40 g of water then, total mass of the solution will be = 70g .

Hence the percentage composition of oleum = (30/70) x 100 = 42.86% .

Hence the percentage composition of water = (40/70) x 100 = 57.14%.

Answer 2

Answer:

Above answer is crct .

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