If 1 litre glucose solution is added to 2 litre of water calculate its molarity
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Correct option is
D
0.18g
Depression in freeing point,
ΔT
f
=
w
1
×M
2
100K
f
w
2
or w
2
=
100K
f
ΔT
f
w
q
M
2
We are given,
w
1
(solvent)=1L=1000g
M
2
(solute)=180gmol
−1
K
f
ΔT
f
=10
−3
w
2
=?
By putting these value in the formula, we get
w
2
=
1000
10
−3
×1000×180
w
2
=0.18g.
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