If P° and P are the vapour pressure of solvent and solution n1 and n2 are the moles of solute and solvent then

Question

If P° and P are the vapour pressure of solvent and solution n1 and n2 are the moles of solute and solvent then (A) P °= P (( n 2/ n 1+ n 2)) (B) P = P °(( n 2/ n 1+ n 2)) (C) P °= P (( n 1/ n 1+ n 2)) (D) P °= P × n 1

Tardigrade Admin · Accepted Answer

P=P0 x P°= solvent P=P0((n2/n1+n2)) x= mole fraction of solvent =( text Moles of solvent / text moles of solute + text moles of solvent ) =(n2/n1+n2)

Q. If $P^{\circ}$ and $P$ are the vapour pressure of solvent and solution $n_{1}$ and $n_{2}$ are the moles of solute and solvent then

$P^{\circ} = P (\frac{n _{2}}{n _{1} + n _{2}})$

$P = P^{\circ} (\frac{n _{2}}{n _{1} + n _{2}})$

$P^{\circ} = P (\frac{n _{1}}{n _{1} + n _{2}})$

$P^{\circ} = P \times n_{1}$

$P = P^{0} x$
$P^{\circ} =$ solvent
$P = P^{0} (\frac{n _{2}}{n _{1} + n _{2}})$
$x =$ mole fraction of solvent $= \frac{Moles of solvent}{moles of solute + moles of solvent}$
$= \frac{n _{2}}{n _{1} + n _{2}}$

Q. If P∘ and P are the vapour pressure of solvent and solution n1​ and n2​ are the moles of solute and solvent then

P∘=P(n1​+n2​n2​​)

P=P∘(n1​+n2​n2​​)

P∘=P(n1​+n2​n1​​)

P∘=P×n1​