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Concentration terms (molarity, molality, mole fraction, ppm) and dilution
JEE Main level
Single correct option
Difficulty 2 of 3
mc078
18
g
of glucose (
180
g/mol
) is dissolved in
250
g
of water. The molality of the solution is
A
0.37
m
B
0.4
m
C
0.1
m
D
0.072
m
Answer: B
Moles of glucose = 18/180 = 0.1. Mass of solvent = 0.250 kg. Molality = 0.1/0.250 = 0.4 mol/kg.
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