The molal lowering of vapour pressure of a liquid is 1.008 mm Hg at 25oC in a very dilute solution containing non volatile solute.The vapour pressure of the liquid at 25oC
is 'A'x10 mm Hg.The value of 'A' is...... [ Integer type ]
Note: Molecular weight of liquid =18.
Steps plz...
21Molal lowering of V.P.=1.008
i.e. if son. is 1 molal conc. then ∂V.P.=1.008
For 1 molar conc:
No of moles of solutes=1.
Wt of soln=1000g
therefore no. of moles of solute=1000/18.
Relative decrease of V.P.=mole fraction of solute
∂V.P./V.P.=1/(1000/18)
V.P.=1.008(1000)/18=56.
A=5.6
Nearest integer=6(if we do not approximate using very dil. son. the ans is closer to 6)
1@Swalraj Sharma, you have done a mistake, it would be
V.P.=1/(1000/18 + 1)
but still the answer comes to be 5.695
