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Question
Chemistry
The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol-1) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75 %, is :
Q. The mass of a non-volatile, non-electrolyte solute (molar mass
=
50
g
m
o
l
−
1
) needed to be dissolved in
114
g
octane to reduce its vapour pressure to
75%
,
is :
7466
198
JEE Main
JEE Main 2018
Solutions
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A
37.5 g
26%
B
75 g
17%
C
150 g
47%
D
50 g
10%
Solution:
Relative lowering of vapour pressure is given by
p
Δ
p
=
x
solute
=
w
A
/
M
A
+
w
a
/
M
B
w
B
/
M
B
where
w
A
and
w
B
are the masses of solvent and solute taken and
M
A
and
M
B
are the molar masses of the solvent and solute.
100
75
=
w
B
/50
+
114/114
w
B
/50
0.75
=
w
B
/50
+
1
w
B
/50
⇒
w
B
=
150
g