If P0 and PS are the vapour pressure of solvent and solution, respectively and N1 and N2 are the moles of solute and solvent then

Question

If P0 and PS are the vapour pressure of solvent and solution, respectively and N1 and N2 are the moles of solute and solvent then (A) (P0-PS/P0)=(N1/N1+N2) (B) (P0-PS/PS)=(N1/N2) (C) PS=P0 ⋅ (N2/N1+N2) (D) All of these.

Tardigrade Admin · Accepted Answer

According to Raoult's law, (P0-PS/P0)=(N1/N1+N2) ; ∴ 1-(PS/P0)=(N1/N1+N2) or (PS/P0)=1-(N1/N1+N2)=(N2/N1+N2) or PS=P0 × (N2/N1+N2) Again, (P0-PS/P0)=(N1/N1+N2) ∴ (P0/P0-PS)=(N1+N2/N1)=1+(N2/N1) or (PS/P0-PS)=(N2/N1) or (P0-PS/PS)=(N1/N2)

Q. If $P_{0}$ and $P_{S}$ are the vapour pressure of solvent and solution, respectively and $N_{1}$ and $N_{2}$ are the moles of solute and solvent then

$\frac{P _{0} - P _{S}}{P _{0}} = \frac{N _{1}}{N _{1} + N _{2}}$

$\frac{P _{0} - P _{S}}{P _{S}} = \frac{N _{1}}{N _{2}}$

$P_{S} = P_{0} \cdot \frac{N _{2}}{N _{1} + N _{2}}$

All of these.

Q. If P0​ and PS​ are the vapour pressure of solvent and solution, respectively and N1​ and N2​ are the moles of solute and solvent then

P0​P0​−PS​​=N1​+N2​N1​​

PS​P0​−PS​​=N2​N1​​

PS​=P0​⋅N1​+N2​N2​​

All of these.

Q. If $P_{0}$ and $P_{S}$ are the vapour pressure of solvent and solution, respectively and $N_{1}$ and $N_{2}$ are the moles of solute and solvent then

$\frac{P _{0} - P _{S}}{P _{0}} = \frac{N _{1}}{N _{1} + N _{2}}$

$\frac{P _{0} - P _{S}}{P _{S}} = \frac{N _{1}}{N _{2}}$

$P_{S} = P_{0} \cdot \frac{N _{2}}{N _{1} + N _{2}}$