Question

The statement(s) true for a 109% oleum sample is/are:

• A. The % of free $S{O}_3$ in sample is 25%.
• B. 13.5g of water should be added to completely hydrolyse 150 of sample.
• C. 150g of sample is sufficient for oxidation of 105.9g of Cu turnings after hydrolysis.
• D. 100g of sample can neutralize 2L of 1.11M of aq. NaOH solution.

Answer 1

Answer: (B), (C), (D)

The correct options are
B

13.5g of water should be added to completely hydrolyse 150 of sample.

C

150g of sample is sufficient for oxidation of 105.9g of Cu turnings after hydrolysis.

D

100g of sample can neutralize 2L of 1.11M of aq. NaOH solution.

Let % of free $S{O}_3$ in sample is x
then, $109=100+\frac{18x}{80}$
x = 40%
The mass of $H_{2}S{O}_4$ formed after hydrolysis $=150\times \frac{109}{100}=163.5g$
Hence mass of water added = (163.5 - 150) = 13.5g
Equivalent of Cu = equivalent of $H_{2}S{O}_4$
w = 105.9gm
Equivalent of NaOH = Equivalent of $H_{2}S{O}_4$
$2\times 1.11\times 1=\frac{w}{48}$
w = 109 gm
Hence oleum required = 100g