wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

0.75 g of oleum sample was diluted with water. If the solution required 62.5 mL of 0.25 N NaOH, the % of free SO3 in the sample will be:

A
9.3%
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
23.5%
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
26.0%
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
12.5%
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 9.3%
Let x g of SO3 is present in the oleum sample, then the amount of H2SO4 will be (0.75x) g.

Equivalents of H2SO4+ Equivalents of SO3= Equivalents of NaOH
0.75x98×2+x80×2=0.25×62.51000
x0.07 g
% of free SO3=0.070.75×100%
% of free SO39.26%

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Introduction
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon