Question

Explain the Raoult's law briefly?
for (5 mark question)​

Answer 1

Answer:

★It state that in the case of a solution of volatile liquid the particle vapour pressure of each component of the solution is directly proportional to its mole fraction

★Let as consider a solution of A&B must be volatile

• The vapour pressure of pure A=P°A
• The vapour pressure of pure A=P°AThe vapour pressure of pure B=P°B

Check this answer is correct

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.

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

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

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

$\boxed{P_A=k_A.X_A} \\ </p><p> \boxed{P_B=k_B.X_B}$

$\mathtt{1) \: P_A=k_A.X_A}$

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

$\mathsf{P_A^0=k_A×1} \\ \\ </p><p>\mathsf{ P_A^0= k_A} \\ \\ </p><p>\mathsf{∴P_A=P_A^{0}X_A}$

$\boxed{ \star \: \mathtt{It \: is \: similarly \: applicable \: to \: P_B \: </p><p>(i.e) \: P_B= P_B^{0}X_B}}$

$\mathsf{P=P_A+P_B} \\ </p><p>\mathsf{= P_A^{0}X_A+P_B^{0}X_B} \\ </p><p>\mathsf{P=P_A^{0}X_A+P_B^0(1-x_{A})} \\ </p><p>\mathsf{P=P_A^{0}X_A+P_B^0-P_B^{0}X_A} \\ </p><p>\boxed{\mathrm{P=P_B^0+X_A(P_A^0-P_B^0)}}$

★ 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.

$\mathsf{ \underline{Derivation:}}$

$P_S \propto X_A \\ </p><p>Ps = P^0Xa </p><p> \mathtt{where, } \mathsf{P^0= Vapour \: pressure \: of \: pure \: solvent }$

$\mathsf{P_{S}= P^{0}(1-X_B)=P^0-P^{0}_{XB}}</p><p>$

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

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

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