Question

# Percentage of free $$SO_3$$ in an oleum bottle labelled 113.5% $$H_2SO_4$$ is :

A
40
B
60
C
50
D
45

Solution

## The correct option is B 60We have$$H_2S_2O_7 + H_2O \rightarrow 2H_2SO_4$$$$113.5$$%$$H_2SO_4$$ means 100g of oleum conatins 113.5 g of $$H_2SO_4$$so, amount of water reacted with oleum = 113.5 - 100 = 13.5gmoles of water = $$\dfrac{13.5}{18}= 0.75 moles$$we also have,$$H_2O + SO_3 \rightarrow H_2SO_4$$so, 1 mole of water reacts with 1 mole of $$SO_3$$.so, 0.75 moles of water reacts with 0.75 moles of $$SO_3$$so, moles of $$SO_3$$ = 0.75so, amount of $$SO_3$$ = $$moles \times molar\ mass = 0.75 \times 80 = 60g$$Chemistry

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