If P0 and PS are the vapour pressure of solvent and solution respectively and N 1 and N 2 are the moles of solute and solvent then

Question

If P0 and PS are the vapour pressure of solvent and solution respectively and N 1 and N 2 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 ( P 0- P S / P 0)=( N 1/ N 1+ N 2) ; ∴ 1-( P S / P 0)=( N 1/ N 1+ N 2) or ( P S / P 0)=1-( N 1/ N 1+ N 2)=( N 2/ N 1+ N 2) or P S = P 0 × ( N 2/ N 1+ N 2) Also we can derived from ( P 0- P S / P 0)=( N 1/ N 1+ N 2) ∴ ( P 0/ P 0- P S )=( N 1+ N 2/ N 1)=( N 2/ N 1)+1 or ( P S / P 0- P S )=( N 2/ N 1) or ( P 0- P S / P S )=( N 1/ N 2)

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

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

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

$P_{s} = P_{0} \times \frac{N _{2}}{N _{1} + N _{2}}$

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

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

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

$P_{s} = P_{0} \times \frac{N _{2}}{N _{1} + N _{2}}$