Question

Explain the Roult's Law briefly with derivation​

Answer 1

Answer:

Raoult's law:

statement-

"For solution of two volatile liquid the vapour pressure of each liquid in a solution is less than the respect to Vapour pressure of pure liquid and vapour pressure of liquid."

"vapour pressure of liquid is directly proportional to its mole fraction"

pA = kXA

hope it helps you

Answer 2

$\star \: \: \mathrm{Roult's \: law \: states \: that \: in \: the }\: \\ \mathrm{ case \: of \: a \: solution \: of \: volatile } \\ \mathrm{ \: liquids, \: the \: partial \: vapour } \\ \mathrm{ \: pressure \: of \: each \: component \: of \: } \\ \mathrm{ t he \: solution \: is \: directly } \\ \mathrm{\: proportional \: to \: its \: mole \: fraction.}$

Roult's Law for the solution containing Volatile solute in Volatile Solvent:

★ Let us consider a solution of two components as A & B which must be volatile.

$\mathsf{ \bullet The \: vapour \: pressure \: of \: pure \: A = P_A ^{0} }$

$\mathsf{ \bullet The \: vapour \: pressure \:of \: pure \: A =P_B^{0}}$

$\boxed{P_A \propto \: X_A}$

$\boxed{P_A=k_A.X_A----(1)}$

${P_B=k_B.X_B}$

$P_A=k_A.X_A$

$\mathsf{ \bullet \: For \: pure \: A \: component \: P_A = P_A^0; \: X_A=1}$

$\mathsf{P_A^0=k_A×1}$

$\mathsf{=> P_A^0= k_A}$

$\mathsf{∴ \: P_A=P_A^{0}X_A}$

$\mathtt{It \: is \: similarly \: applicable \: to \: P_B}$

${(i.e) P_B= P_B^{0}X_B}$

$\mathsf{P=P_A+P_B}$

$\mathsf{= P_A^{0}X_A+P_B^{0}X_B}$

$\mathsf{P=P_A^{0}X_A+P_B^0(1-x_{A})}$

$\mathsf{P=P_A^{0}X_A+P_B^0-P_B^{0}X_A}$

$\boxed{\mathrm{P=P_B^0+X_A(P_A^0-P_B^0)}}$

Roult's Law for the solution containing Non-Volatile solute in Volatile Solvent:

★ A solution Containing non volatile solute (B) in volatile solvent (A) having the relative lowering of Vapour pressure is equal to mole fraction of the solute.

Derivation:

$P_S \propto X_A$

$P_S = P^{0}X_A$

where,

$\mathsf{P^0= \: Vapour \: pressure \: of \: pure \: solvent }$

$\mathsf{P_S= P^{0}(1-X_B)=P^0-P^0_{XB}}$

$\mathsf{P^S+P_{x_B}^0= P^0}$

$\mathsf{P_{x_B}^0= P^0-P^s}$

$P^0-P^s$

$\boxed{X_B= \frac{P^0-P^s}{P^0} }$