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Concentration terms (molarity, molality, mole fraction, ppm) and dilution
JEE Advanced level
Single correct option
Difficulty 3 of 3
mc160
A commercial nitric acid is 68%
HNO
X
3
by mass with density
1.41
g
m
L
−
1
(
M
=
63
). The volume of this acid needed to prepare 500 mL of 0.50 M
HNO
X
3
is
A
24.2
mL
B
32.9
mL
C
16.4
mL
D
11.2
mL
Answer: C
Molarity of the acid = 1.41 x 1000 x 0.68/63 = 15.22 M. Moles needed = 0.500 x 0.50 = 0.25 mol. Volume = 0.25/15.22 = 0.0164 L = 16.4 mL.
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