what is the correct formula to find relative lowering of vapor pressure on adding solute according to RAOULTS law
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according to Raoults law the relative lowering of vapour pessure is equal to the mole fraction of solute
V. P Of pure solvent = p o
Relative of soultion =p s
Lowering in V. P = po -ps
Relative lowering in V. P = po-ps / po
Number of moles of. Solute = n
No. Of moles of solvent = N
Therefore total no. Of moles = n+N
Mole fraction Of solute = n/ n+N
According to Raoult's law
Po-ps / po = n / n+ N
V. P Of pure solvent = p o
Relative of soultion =p s
Lowering in V. P = po -ps
Relative lowering in V. P = po-ps / po
Number of moles of. Solute = n
No. Of moles of solvent = N
Therefore total no. Of moles = n+N
Mole fraction Of solute = n/ n+N
According to Raoult's law
Po-ps / po = n / n+ N
Answered by
1
Raout's law:
relative lowering of vapour pressure = mole fraction of solute
(P₀ - P)/P₀ = x / (s + x)
P₀ = partial vapour pressure of the solvent (pure)
P = partial vapour pressure of the solvent with some solute (x moles)
s = number of moles of solute inside solvent
relative lowering of vapour pressure = mole fraction of solute
(P₀ - P)/P₀ = x / (s + x)
P₀ = partial vapour pressure of the solvent (pure)
P = partial vapour pressure of the solvent with some solute (x moles)
s = number of moles of solute inside solvent
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