The relative lowering of vapour pressure of a solution of 6g of urea in 90g of water is equal to
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According to Raoult’s law, the relative lowering of Vapour pressure is equal to mole fraction of solute.
Δp/p = Mole fraction of solute
= n / (n + n’)
= (6 g / 60 g/mol) / [(6 g / 60 g/mol) + (90 g / 18 g/mol)]
≈ 0.02
[Here 60 g/mol is Molar mass of urea and 18 g/mol is Molar mass of water]
Relative lowering of Vapour pressure is 0.2
Refer to the image attched for the better understanding.
Δp/p = Mole fraction of solute
= n / (n + n’)
= (6 g / 60 g/mol) / [(6 g / 60 g/mol) + (90 g / 18 g/mol)]
≈ 0.02
[Here 60 g/mol is Molar mass of urea and 18 g/mol is Molar mass of water]
Relative lowering of Vapour pressure is 0.2
Refer to the image attched for the better understanding.
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RLVP formula
w1/m1×m2/w2
w2=90g
m2=18g
w1=6g
m1=molecular weight of urea and substitute in that formula
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