Mole ConceptHard

Question

In 200 g of a sample of oleum labelled as 109.0%, 12 g water is added. The new labelling of the oleum sample is

Options

A.106.0%
B.103.0%
C.102.8%
D.105.6%

Solution

200 g of 109% oleum requires 18 g of water for complete conversion into H2SO4. On adding 12 g of water, the resulting 212 g oleum is still requiring 6 g of water. Hence, the new labelling is

$$\left( 100 + \frac{6}{212} \times 100 \right) = 102.83\%$$

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