Question

If in a sample of oleum, mole fraction of so3 is 0.5 , lable the oleum sample?

Answer 1

Given:

Mole fraction of SO3, X = 0.5

To Find:

Label the oleum sample.

Calculation:

- The reaction for conversion of SO3 to H2SO4 is:

SO3 + H2O → H2SO4

⇒ 1 mole SO3 requires 1 mole of H2O.

⇒ 0.5 mole SO3 requires 0.5 mole of H2O.

- Mass of 0.5 moles of water = 0.5 × 18 = 9 gm

- Oleum concentration = (100 + mass of water) = 100 + 9

⇒ Oleum concentration = 109 %

- Hence, the oleum sample is labelled as 109% oleum.

Answer 2

Answer:

110.11%

Explanation:

Mole fraction of SO3 = 0.5

H20          + SO3      --> H2SO4

0.5 mole   0.5 mole      0.5 mole

mass of SO3 = 0.5×80 = 40gm

mass of H2SO4 = 0.5×98 = 49gm

mass of solution = 40+49 = 89gm

mass of H2O = 0.5×18 = 9gm

89gm solution contain 9gm H20

100gm solution contain H2O = (9÷89)×100 =10.11gm

% labelling of oleum sample = ( 100+10.11 )% = 110.11%.