The vapor pressure of 100 g of water reduces from 17.53 mm to 17.22 mm Hg when 17.10 g of non-electrolyte substance X is dissolved in it. Substance X can be
Answers
Answer:
x is glucose
Explanation:
12th
Chemistry
Solutions
Vapour Pressure of Liquid Solutions
The vapour pressure of 100 ...
CHEMISTRY
The vapour pressure of 100 g water reduces from 3000 Nm
−2
to 2985 Nm
−2
when 5g of substance X is dissolved in it. The substance X is :
October 15, 2019
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Shamil Sisangia
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VIDEO EXPLANATION
ANSWER
Raoult's law states that relative lowering of vapour pressure and is equal to
the mole fraction of the solute.
p
1
p
1
0
−p
1
=x
2
p
1
0
=3000 Nm
−2
= vapour pressure of solution.
p
1
=2985 Nm
−2
= vapour pressure of sollvent.
x
2
= mole-fraction of solute.
Let molecular wt. of X is x.
No of moles of X is
x
5
No of moles of solvent is
18
100
.
Mole-fraction of X is
(100/18)
(5/x)
=9/10x
So,
3000
3000−2985
=
10
9
x
⟹x=180
The molecular weight of glucose is 180 g/mol.
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