Question

A sample of Oleum is labelled 118% . The percentage of free SO3 in the sample is

Answer 1

The percentage of free $SO_{3}$ in the sample is 80%.

Explanation:

• Oleum as we know it is a mixture of both $so_{3}$  as well as $H_{2}SO_{4}.$
• Hence, if percentage of the labelled concentration is given as 118%.
• Then it shall have certain amount of $so_{3}$ present in it now the amount of $So_{3}$ can be found out using the formula

                           $\frac{98x}{80} + (100-x )= 118$

• Here we know the moles of $H_{2}So_{4}$ is given as 98.
• Hence in the formula On calculating the above given formula we get the value of X= 80 .
• So the percent of $So_{3}$ present in the mixture would be $\frac{80}{100} \times 100= 80%.$

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