A 6.85 g sample of the hydrate Sr(OH)2.xH2O is dried in an oven to give 3.13 g of anhydrous Sr(OH)2. what is the value of x ? (Atomic weights : Sr = 87.60, O = 16.0, H = 1.0)
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
8
b
12
c
10
d
6
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Sr(OH)26.85 g⋅xH2O⟶Sr(OH)23.13 g+xH2O3.72 g% of water =3.726.85×100=18x121+18x×100 ⇒x=8