Chemical Engineering - Stoichiometry - Discussion
Discussion Forum : Stoichiometry - Section 4 (Q.No. 40)
40.
A solution of specific gravity 1 consists of 35% A by weight and the remaining B. If the specific gravity of A is 0.7, the specific gravity of Bis
Discussion:
1 comments Page 1 of 1.
AAMIR said:
7 years ago
Basis =1 kg of solution.
wt.of A= 0.35 kg.
wt.of B=0.65,
sp.gravity= (wt.of solution)/(volume of the solution),
1=0.35+0.65/((0.35/0.7)+(0.65/x)(sp.gravity of B)).
0.350/70+0.65/x=1.
0.65/x=1/2 * X=2 * 0.65 = 1.3.
wt.of A= 0.35 kg.
wt.of B=0.65,
sp.gravity= (wt.of solution)/(volume of the solution),
1=0.35+0.65/((0.35/0.7)+(0.65/x)(sp.gravity of B)).
0.350/70+0.65/x=1.
0.65/x=1/2 * X=2 * 0.65 = 1.3.
(4)
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