Solutions MCQ Questions & Answers in Physical Chemistry | Chemistry

At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers.

$$\eqalign{ & {\text{Normality}} \cr & = \frac{{{\text{Weight percentage }} \times {\text{ density }} \times {\text{10}}}}{{{\text{Equivalent weight}}}} \cr & = \frac{{98 \times 1.8 \times 10}}{{49}} \cr & = 36\,N \cr & {N_2} = 2 \times 0.1\,N \cr & \,\,\,\,\,\,\,\,\, = 0.2\,N \cr & \,\,\,\,{N_1}{V_1} = {N_2}{V_2} \cr & 36 \times V = 0.2 \times 1000 \cr & \,\,\,\,\,\,\,\,\,\,\,V = \frac{{0.2 \times 1000}}{{36}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5.55\,mL\left( {{\text{dissociation of acid}}} \right) \cr} $$

$$CaC{l_2}$$  is an electrolyte and dissociates to give three ions as
$$CaC{l_2} \to C{a^{2 + }} + 2C{l^ - }$$
While $$KCl$$  gives two ions and glucose and urea are non-electrolytes, so remains undissociated.
As osmotic pressure is a colligative property ( i.e. depend only on number of particles ), so highest for $$CaC{l_2}.$$

Reverse osmosis The minimum external pressure applied to a solution separated from a solvent by semipermeable membrane to prevent osmosis, is called osmotic pressure. When the pressure applied to solution is more than osmotic pressure, solute will pass from the solution into solvent through the semipermeable membrane. This phenomenon is known as reverse osmosis.
The osmotic pressure of sea water is $$25$$ $$atm$$  at $${15^ \circ }C.$$  When pressure greater than $$26$$ $$atm$$  is applied on sea water separated by a rigid semipermeable membrane, pure water is obtained. This is also called desalination of sea water.

$${H_2}O\left( {43\% } \right)$$   and $$HI\left( {57\% } \right)$$   make maximum boiling azeotropic mixture on boiling at $$400\,K.$$

$$\eqalign{ & P_{{\text{solvent}}}^ \circ = 760\,torr\,\left( {{\text{at}}\,{\text{the}}\,b.pt.} \right); \cr & {n_{{\text{solute}}}} = 1 \cr & {n_{{\text{solvent}}}} = \frac{{1000}}{{40}} = 25; \cr & \Delta P = 760 \times {X_{{\text{solute}}}} \cr & = \frac{{760}}{{26}}torr \cr & = 29.23\,torr \cr} $$

$$C{H_3}COC{H_3} + CHC{l_3}$$     is an example of solution showing negative deviation from Raoult's law since $$A - B$$  attractions ( acetone + chloroform ) are more than $$A-A$$  ( acetone - acetone ) and $$B - B$$  ( chloroform - chloroform ) attractions.

$$\eqalign{ & \because \,\,\Delta {T_b}^ \circ = {T_b} - {T_b}^ \circ \cr & {\text{Where }}{{\text{T}}_b}{\text{ = b}}{\text{.pt of solution}} \cr & {\text{T}}_b^ \circ \, = {\text{b}}{\text{.pt of solvent or}}\,\,\,{{\text{T}}_b} = {T_b}^ \circ = T_b^ \circ + \Delta {T_b} \cr} $$
NOTE : Elevation in boiling point is a colligative property, which depends upon the no. of particles. Thus greater the number of particles, greater is it elevation and hence greater will be its boiling point.
$$N{a_2}S{O_4} \rightleftharpoons 2\mathop {Na}\limits^ + \, + SO_4^{2 - }$$
Since $$N{a_2}S{O_4}$$  has maximum number of particles (3) hence has maximum boiling point.

$$\eqalign{ & {\text{Number of particles (}}i{\text{)}} \cr & \left( 1 \right)\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]C{l_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \cr & \left( 2 \right)\left[ {Co{{\left( {{H_2}O} \right)}_5}Cl} \right]C{l_2}.{H_2}O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3 \cr & \left( 3 \right)\left[ {Co{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]Cl.2{H_2}O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \cr & \left( 4 \right)\left[ {Co{{\left( {{H_2}O} \right)}_3}C{l_3}} \right].3{H_2}O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4 \cr & \Delta {T_f} \propto i\,;\,{\text{where }}\Delta {{\text{T}}_f} = \left( {{T_f} - T{'_f}} \right) \cr} $$
Remember, the greater the no. of particles, the lower will be the freezing point. Compound (D) will have the highest freezing point due to least no of particles.

$$\eqalign{ & {\text{Given, molality,}}\,\,m = 0.0020\,m \cr & \Delta {T_f} = {0^ \circ }C - {0.00732^ \circ }C \cr & \,\,\,\,\,\,\,\,\,\,\,\, = - {0.00732^ \circ }C \cr & \,\,\,\,\,{k_f} = - {1.86^ \circ }C/m \cr & \Delta {T_f} = i \cdot {k_f} \times m \cr & \,\,\,\,\,\,\,\,i = \frac{{\Delta {T_f}}}{{{k_f} \times m}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{0.00732}}{{1.86 \times 0.0020}} \cr & \,\,\,\,\,\,\,\,\,\,\, = 1.96 \approx 2 \cr} $$
Since, the compound is ionic, so number of moles produced is equal to van't Hoff factor, $$i.$$
Hence, $$2\,moles$$  of ions are produced.
$$\mathop {\left[ {Co{{\left( {N{H_3}} \right)}_5}N{O_2}} \right]}\limits_{1\,mol} Cl \to \underbrace {{{\left[ {Co{{\left( {N{H_3}} \right)}_5}N{O_2}} \right]}^ + } + C{l^ - }}_{2\,ions}$$