Chemical Engineering - Stoichiometry - Discussion
Discussion Forum : Stoichiometry - Section 3 (Q.No. 29)
29.
An aqueous solution of 2.45% by weight H2SO4 has a specific gravity of 1.011. The composition expressed in normality is
Discussion:
1 comments Page 1 of 1.
Jay said:
6 years ago
Molarity={(W/W%)*10*1.011} ÷ {Molecular weight of solute}.
= {2.45*10*1.011}÷{98}.
= 0.25275.
Now,
Normality = Molarity * n factor
= 0.25275 * 2,
= 0.5055,
(For H2SO4 n factor is 2).
= {2.45*10*1.011}÷{98}.
= 0.25275.
Now,
Normality = Molarity * n factor
= 0.25275 * 2,
= 0.5055,
(For H2SO4 n factor is 2).
(3)
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