Question

Label an oleum sample which has mass fractions of h2so4 as 0.8

Answer 1

Answer:

104.5% labelling.

Explanation:

Please see image for Reason.

Answer 2

Final answer: % label of oleum sample = 104.5

Given that: We are given mass fractions of $H_{2}SO_{4}$ is 0.8.

To find: We have to find label an oleum sample which has mass fractions of  $H_{2}SO_{4}$ as 0.8.

Explanation:

• Oleum also known as fuming sulfuric acid is produce by contact process. Sulfuric acid is manufactured by contact process in which oleum formation is one step in this process.
• The chemical formula is  $H_{2}S_{2} O_{7}$.

• In order to get desired concentration $H_{2}SO_{4}$ by diluting oleum with water.

• $SO_{3} +H_{2}SO_{4}$ → $H_{2}S_{2} O_{7}$(oleum)

• $H_{2}S_{2} O_{7}+H_{2}O$ → $H_{2}SO_{4}$

• We can write $SO_{3} +H_{2}SO_{4}$ → $H_{2}SO_{4}$ in the above chemical equation.

$H_{2}SO_{4}+SO_{3} +H_{2}O$ → $H_{2}SO_{4}$

• Given that mass fraction $H_{2}SO_{4}$= 0.8

        Which means 80 g $H_{2}SO_{4}$  and 20 g of $SO_{3}$

• $Number of moles = \frac{Given mass}{Molecular mass}$

• $Number of moles of SO_{3} = \frac{Given mass of SO_{3}}{Molecular mass of SO_{3}}$

• $Given mass of SO_{3} = 20 g$

• $Molecular mass of SO_{3} = 80g$

• $Number of moles of SO_{3} =\frac{20}{80} =\frac{1}{4} mole$

• Also,

$Moles of SO_{3} = Moles of H_{2}O =\frac{1}{4}$

$Moles of H_{2}O = \frac{label - 100}{18}\\\\\frac{1}{4} = \frac{label - 100}{18} \\\\label - 100 =\frac{18}{4} = \frac{9}{2} \\\\label = 100 +\frac{9}{2}\\\\label = \frac{200+9}{2} =\frac{209}{2}=104.5$

• Hence, % label of oleum sample = 104.5

To know more about the concept please go through the links

https://brainly.in/question/16990597

https://brainly.in/question/16120081

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