According to Raoult's law, the relative lowering in vapour pressure of a solution is equal to

Question

According to Raoult's law, the relative lowering in vapour pressure of a solution is equal to (A) moles of solute (B) mole fraction of solvent (C) moles of solvent (D) mole fraction of solute

Tardigrade Admin · Accepted Answer

According to Raoulfs law, the relative lowering of vapour pressure of solution is equal to mole fraction of solute.

\frac{p - p _{s}}{p _{s}} = X_{B}

where,

p =

Vapour pressure of pure solvent

P_{s} =

Vapour pressure of solution

X_{B} =

Mole fraction of solute

Q. According to Raoult's law, the relative lowering in vapour pressure of a solution is equal to

moles of solute

mole fraction of solvent

moles of solvent

mole fraction of solute

According to Raoulfs law, the relative lowering of vapour pressure of solution is equal to mole fraction of solute. ps​p−ps​​=XB​ where, p= Vapour pressure of pure solvent Ps​= Vapour pressure of solution XB​= Mole fraction of solute

According to Raoulfs law, the relative lowering of vapour pressure of solution is equal to mole fraction of solute.
$\frac{p - p _{s}}{p _{s}} = X_{B}$
where, $p =$ Vapour pressure of pure solvent
$P_{s} =$ Vapour pressure of solution $X_{B} =$
Mole fraction of solute