A solution is prepared by dissolving 5g of non volatile solute in 95 g of water. It has a vapour pressure of 23.375 mm hg at 25°c.Calculate the molar mass of the solute. (vp of pure water at 25°c is 23.75mm hg)
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Answered by
35
Answer:
Explanation:
Rlvp= mole fraction of solute
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Answered by
19
Answer:
Explanation:
Quantity of water = 95g (Given)
Vapour pressure = 23.375mm (Given)
Temperature = 25°c (Given)
According to the Raoult’s Law -
p0 - p/p0 = w × M/ m× W
where -
p0 is the vapour pressure of pure water
p is the vapour pressure of solvent
w is the weight of solute
m is the molecular weight of solute
W is the molecular weight of solvent
M is the Molar Mass of Water = 18 g mol -1
Substituting the values in the formula, we will get -
= 23.75 - 23.375 / 23.75 = 5 × 18/ m × 95
= 23.75 - 23.375 / 23.75 = 90/95m
= 60m/s
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