Question

200 gram sample of 122.5% oleum is diluted with distilled water. How many mole of sodium
hydroxide would be required to neutralize completely the diluted oleum.
80 points.....please...I want just the answer... pleeeaaaassse​

Answer 1

Explanation:

ANSWER

Oleum consists of SO

3

and H

2

SO

4

.

Let the mass of SO

3

in the given sample of oleum be =x g

Mass of H

2

SO

4

in the given sample of oleum =(0.5−x)g

Eq. mass of SO

3

=

2

80

=40

No. of g equivalents of SO

3

=

40

x

[2NaOH+SO

3

→Na

2

SO

4

+H

2

O

2NaOH+H

2

SO

4

→Na

2

SO

4

+2H

2

O]

Eq. mass of H

2

SO

4

=

2

98

=49

No. of g equivalents of H

2

SO

4

=

49

(0.5−x)

Total no. of g equivalents =

40

x

+

49

(0.5−x)

26.7 mL of 0.4 N NaOH contain no. of equivalents of NaOH

=

1000

0.4

×26.7

At equivalence point,

No. of g equivalents of NaOH=

40

x

+

49

(0.5−x)

So,

1000

0.4×26.7

=

40×49

49x+(40×0.5−40x)

x=

9

0.9328

=0.1036

Hence, % of free SO

3

=

0.5

0.1036

×100

=20.72%

Answer 2

⭐⭐⭐Answer⭐⭐⭐

The number of equivalents of H2​SO4​ is equal to the number of equivalents of SO3​. It is also equal to the number of equivalents of NaOH.

Hence, 98x​×2+80(1−x)×2​=54×0.4×10−3

or, x=0.74

The percentage of free SO3​=11−0.74​×100=26%