12 gm of a solid solute dissolved in 90 gm pure water .vapour pressure of the resulting solution is 750mm-hg at 100°c.calucate molecular mass of the solute . [Solute does not undergo any dissociation or association in its aqueous solution]
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Explanation:
ANSWER
According to raoult's law,
p
0
p
0
−p
=
m×W
w×M
Where,
p
0
is Vapour pressure of pure water
p is vapour pressure of solvent
w is weight of solute
m is molecular weight of solute
W is molecular weight of solvent
32
32−31.84
=
m×200
10×18
⇒m=
0.005
0.9
=180
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