319030
The correct relationship between the boiling points of very dilute solutions of \({\rm{AlC}}{{\rm{l}}_{\rm{3}}}\left( {{{\rm{t}}_{\rm{1}}}} \right)\,\,{\rm{and}}\,\,{\rm{CaC}}{{\rm{l}}_{\rm{2}}}\left( {{{\rm{t}}_{\rm{2}}}} \right)\), having the same molar concentration, is
More is the number of ions higher is the boiling point of solution. \({\rm{AlC}}{{\rm{l}}_{\rm{3}}}\) produces 4 ions and \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) produces 3 ions. So, \({\rm{AlC}}{{\rm{l}}_{\rm{3}}}\) solution possess higher boiling point.
CHXII02:SOLUTIONS
319031
If the van’t Hoff-factor for 0.1 M Ba \({\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}}\) solution is 2.74, the degree of dissociation is
1 0.87
2 0.74
3 0.91
4 87
Explanation:
Given, Molarity = 0.1 M vant Hoff factor (i) = 2.74 Since, \({\rm{i > 1}}\), it means solute is undergoing dissociation. \(Ba{\left( {N{O_3}} \right)_2} \rightleftharpoons B{A^{2 + }} + 2NO_3^ - \) Number of particles dissociated (n) = 3 Now, \({\rm{\alpha }}\) (degree of dissociation) \({\rm{ = }}\frac{{\left( {{\rm{i - 1}}} \right)}}{{\left( {{\rm{n - 1}}} \right)}}\) \({\rm{ = }}\frac{{\left( {{\rm{2}}{\rm{.74 - 1}}} \right)}}{{\left( {{\rm{3 - 1}}} \right)}}\) \({\rm{ = 0}}{\rm{.87}}\)
CHXII02:SOLUTIONS
319032
The freezing point of \(1 \%\) solution of lead nitrate in water will be
1 \(2{ }^{\circ} \mathrm{C}\)
2 \(1^{\circ} \mathrm{C}\)
3 \(0^{\circ} \mathrm{C}\)
4 below \(0^{\circ} \mathrm{C}\)
Explanation:
Aqueous solution of any substance (non-volatile) freezes below \(0^\circ {\rm{C}}\) because the vapour pressure of the solution becomes lower than that of pure solvent.
319030
The correct relationship between the boiling points of very dilute solutions of \({\rm{AlC}}{{\rm{l}}_{\rm{3}}}\left( {{{\rm{t}}_{\rm{1}}}} \right)\,\,{\rm{and}}\,\,{\rm{CaC}}{{\rm{l}}_{\rm{2}}}\left( {{{\rm{t}}_{\rm{2}}}} \right)\), having the same molar concentration, is
More is the number of ions higher is the boiling point of solution. \({\rm{AlC}}{{\rm{l}}_{\rm{3}}}\) produces 4 ions and \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) produces 3 ions. So, \({\rm{AlC}}{{\rm{l}}_{\rm{3}}}\) solution possess higher boiling point.
CHXII02:SOLUTIONS
319031
If the van’t Hoff-factor for 0.1 M Ba \({\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}}\) solution is 2.74, the degree of dissociation is
1 0.87
2 0.74
3 0.91
4 87
Explanation:
Given, Molarity = 0.1 M vant Hoff factor (i) = 2.74 Since, \({\rm{i > 1}}\), it means solute is undergoing dissociation. \(Ba{\left( {N{O_3}} \right)_2} \rightleftharpoons B{A^{2 + }} + 2NO_3^ - \) Number of particles dissociated (n) = 3 Now, \({\rm{\alpha }}\) (degree of dissociation) \({\rm{ = }}\frac{{\left( {{\rm{i - 1}}} \right)}}{{\left( {{\rm{n - 1}}} \right)}}\) \({\rm{ = }}\frac{{\left( {{\rm{2}}{\rm{.74 - 1}}} \right)}}{{\left( {{\rm{3 - 1}}} \right)}}\) \({\rm{ = 0}}{\rm{.87}}\)
CHXII02:SOLUTIONS
319032
The freezing point of \(1 \%\) solution of lead nitrate in water will be
1 \(2{ }^{\circ} \mathrm{C}\)
2 \(1^{\circ} \mathrm{C}\)
3 \(0^{\circ} \mathrm{C}\)
4 below \(0^{\circ} \mathrm{C}\)
Explanation:
Aqueous solution of any substance (non-volatile) freezes below \(0^\circ {\rm{C}}\) because the vapour pressure of the solution becomes lower than that of pure solvent.
319030
The correct relationship between the boiling points of very dilute solutions of \({\rm{AlC}}{{\rm{l}}_{\rm{3}}}\left( {{{\rm{t}}_{\rm{1}}}} \right)\,\,{\rm{and}}\,\,{\rm{CaC}}{{\rm{l}}_{\rm{2}}}\left( {{{\rm{t}}_{\rm{2}}}} \right)\), having the same molar concentration, is
More is the number of ions higher is the boiling point of solution. \({\rm{AlC}}{{\rm{l}}_{\rm{3}}}\) produces 4 ions and \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) produces 3 ions. So, \({\rm{AlC}}{{\rm{l}}_{\rm{3}}}\) solution possess higher boiling point.
CHXII02:SOLUTIONS
319031
If the van’t Hoff-factor for 0.1 M Ba \({\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}}\) solution is 2.74, the degree of dissociation is
1 0.87
2 0.74
3 0.91
4 87
Explanation:
Given, Molarity = 0.1 M vant Hoff factor (i) = 2.74 Since, \({\rm{i > 1}}\), it means solute is undergoing dissociation. \(Ba{\left( {N{O_3}} \right)_2} \rightleftharpoons B{A^{2 + }} + 2NO_3^ - \) Number of particles dissociated (n) = 3 Now, \({\rm{\alpha }}\) (degree of dissociation) \({\rm{ = }}\frac{{\left( {{\rm{i - 1}}} \right)}}{{\left( {{\rm{n - 1}}} \right)}}\) \({\rm{ = }}\frac{{\left( {{\rm{2}}{\rm{.74 - 1}}} \right)}}{{\left( {{\rm{3 - 1}}} \right)}}\) \({\rm{ = 0}}{\rm{.87}}\)
CHXII02:SOLUTIONS
319032
The freezing point of \(1 \%\) solution of lead nitrate in water will be
1 \(2{ }^{\circ} \mathrm{C}\)
2 \(1^{\circ} \mathrm{C}\)
3 \(0^{\circ} \mathrm{C}\)
4 below \(0^{\circ} \mathrm{C}\)
Explanation:
Aqueous solution of any substance (non-volatile) freezes below \(0^\circ {\rm{C}}\) because the vapour pressure of the solution becomes lower than that of pure solvent.
319030
The correct relationship between the boiling points of very dilute solutions of \({\rm{AlC}}{{\rm{l}}_{\rm{3}}}\left( {{{\rm{t}}_{\rm{1}}}} \right)\,\,{\rm{and}}\,\,{\rm{CaC}}{{\rm{l}}_{\rm{2}}}\left( {{{\rm{t}}_{\rm{2}}}} \right)\), having the same molar concentration, is
More is the number of ions higher is the boiling point of solution. \({\rm{AlC}}{{\rm{l}}_{\rm{3}}}\) produces 4 ions and \({\rm{CaC}}{{\rm{l}}_{\rm{2}}}\) produces 3 ions. So, \({\rm{AlC}}{{\rm{l}}_{\rm{3}}}\) solution possess higher boiling point.
CHXII02:SOLUTIONS
319031
If the van’t Hoff-factor for 0.1 M Ba \({\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}}\) solution is 2.74, the degree of dissociation is
1 0.87
2 0.74
3 0.91
4 87
Explanation:
Given, Molarity = 0.1 M vant Hoff factor (i) = 2.74 Since, \({\rm{i > 1}}\), it means solute is undergoing dissociation. \(Ba{\left( {N{O_3}} \right)_2} \rightleftharpoons B{A^{2 + }} + 2NO_3^ - \) Number of particles dissociated (n) = 3 Now, \({\rm{\alpha }}\) (degree of dissociation) \({\rm{ = }}\frac{{\left( {{\rm{i - 1}}} \right)}}{{\left( {{\rm{n - 1}}} \right)}}\) \({\rm{ = }}\frac{{\left( {{\rm{2}}{\rm{.74 - 1}}} \right)}}{{\left( {{\rm{3 - 1}}} \right)}}\) \({\rm{ = 0}}{\rm{.87}}\)
CHXII02:SOLUTIONS
319032
The freezing point of \(1 \%\) solution of lead nitrate in water will be
1 \(2{ }^{\circ} \mathrm{C}\)
2 \(1^{\circ} \mathrm{C}\)
3 \(0^{\circ} \mathrm{C}\)
4 below \(0^{\circ} \mathrm{C}\)
Explanation:
Aqueous solution of any substance (non-volatile) freezes below \(0^\circ {\rm{C}}\) because the vapour pressure of the solution becomes lower than that of pure solvent.