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Concentration terms (molarity, molality, mole fraction, ppm) and dilution
JEE Main level
Single correct option
Difficulty 2 of 3
mc136
The molality of a solution prepared by dissolving 36 g of glucose (
M
=
180
) in 400 g of water is
A
0.46
m
B
0.50
m
C
0.20
m
D
0.00050
m
Answer: B
Moles of glucose = 36/180 = 0.2. Mass of solvent = 0.400 kg. Molality = 0.2/0.400 = 0.50 mol/kg.
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