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Oleum is considered as a solution of $$SO_3$$ in $$H_2SO_4$$, which is obtained by passing $$SO_3$$ in solution of $$H_2SO_4$$.When 100 g sample of oleum is diluted with desired weight of $$H_2O$$ then the total mass of $$H_2SO_4$$ obtained after dilution is known as percentage labelling of oleum. For example , a oleum lablled as $$109\%$$ $$H_2SO_4$$ means 109g total mass of pure $$H_2SO4$$ will be formed when 100g of oleum is diluted by 9 g of $$H_2O$$ which combined with all the free $$SO_3$$ to form $$H_2SO_4$$ as $$SO_3 +H_2O \rightarrow H_2SO_4$$.
1 g of oleum sample is diluted with water. The solution require 54 ml of 0.4 N NaOH for completed neutralization for the % of free $$SO_3$$ in the sample is:


A
74
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B
26
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C
20
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D
None of these
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Solution

The correct option is B 26
equivalent of $$H_2SO_4$$ + equivalent of $$SO_3$$= equivalent of NaOH

$$\dfrac {x}{98} \times 2 + \dfrac {(1-x)\times 2}{80} =54 \times 0.4 \times 10^{-3}$$

x=0.74

% of free $$SO_3= \dfrac {1-0.74}{1}\times 100 =26$$%

Chemistry

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