277326
The latest heat of vaporization of a liquid at $227^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure is $12 \mathrm{kcal} \mathrm{mol}^{-1}$. The change in internal energy, if 3 moles of the liquid changes to vapour at same conditions is
277370
The aqueous solution that has the lowest vapour pressure at a given temperature is
1 0.1 molal sodium phosphate
2 0.1 molal barium chloride
3 0.1 molal sodium chloride
4 0.1 molal glucose
Explanation:
The same concentrations of relative lowering of vapour pressure depends upon the number of ions or particles present in different solutions. $\mathrm{Na}_{3} \mathrm{PO}_{4} \rightarrow 3 \mathrm{Na}^{+}+\mathrm{PO}_{4}^{3-}=4$ ions $\mathrm{BaCl}_{2} \rightarrow \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-}=3$ ions $\mathrm{NaCl} \rightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}=2$ ions Glucose $\rightarrow$ no dissociation $=1$ ions As $\mathrm{Na}_{3} \mathrm{PO}_{4}$ gives maximum ions (4) on dissociation, hence relative lowering of vapour pressure is maximum for sodium phosphate $\left(\mathrm{Na}_{3} \mathrm{PO}_{4}\right)$ and it has lowest vapour pressure.
UP CPMT-2013
SOLUTIONS
277329
If $x_{1}$ and $x_{2}$ represent the mole fraction of a component $A$ in the vapour phase and liquid mixture respectively and $p_{A}^{0}$ and $p_{B}^{0}$ represent Vapour pressure of pure $A$ and $B$, then total of vapour pressure of liquid mixture is
According to Dalton's law of partial pressures of gasses- $\begin{aligned} & \frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\text {total }}}=\mathrm{x}_{1} \\ & \mathrm{P}_{\mathrm{A}}=\mathrm{x}_{1} \times \mathrm{P}_{\text {total }} \quad \ldots \ldots . . \text {. (i) } \end{aligned}$ According to Raoult's law $\mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{x}_{2}$ From equation (i) and equation (ii) we get:- $\begin{aligned} & \mathrm{P}_{\text {total }} \mathrm{x}_{1}=\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{x}_{2} \\ & \mathrm{P}_{\text {total }}=\frac{\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{x}_{2}}{\mathrm{x}_{1}} \end{aligned}$
AMU-2018
SOLUTIONS
277330
A $0.0020 \mathrm{~m}$ aqueous solution of an ionic compound freezes at $-0.00732^{\circ} \mathrm{C}$. Number of moles of in which one mole of the ionic compound produces on being dissolved in water will be $\left(K_{f}=1.88 \mathrm{~K} \mathrm{kgm}^{-1}\right)$
277326
The latest heat of vaporization of a liquid at $227^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure is $12 \mathrm{kcal} \mathrm{mol}^{-1}$. The change in internal energy, if 3 moles of the liquid changes to vapour at same conditions is
277370
The aqueous solution that has the lowest vapour pressure at a given temperature is
1 0.1 molal sodium phosphate
2 0.1 molal barium chloride
3 0.1 molal sodium chloride
4 0.1 molal glucose
Explanation:
The same concentrations of relative lowering of vapour pressure depends upon the number of ions or particles present in different solutions. $\mathrm{Na}_{3} \mathrm{PO}_{4} \rightarrow 3 \mathrm{Na}^{+}+\mathrm{PO}_{4}^{3-}=4$ ions $\mathrm{BaCl}_{2} \rightarrow \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-}=3$ ions $\mathrm{NaCl} \rightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}=2$ ions Glucose $\rightarrow$ no dissociation $=1$ ions As $\mathrm{Na}_{3} \mathrm{PO}_{4}$ gives maximum ions (4) on dissociation, hence relative lowering of vapour pressure is maximum for sodium phosphate $\left(\mathrm{Na}_{3} \mathrm{PO}_{4}\right)$ and it has lowest vapour pressure.
UP CPMT-2013
SOLUTIONS
277329
If $x_{1}$ and $x_{2}$ represent the mole fraction of a component $A$ in the vapour phase and liquid mixture respectively and $p_{A}^{0}$ and $p_{B}^{0}$ represent Vapour pressure of pure $A$ and $B$, then total of vapour pressure of liquid mixture is
According to Dalton's law of partial pressures of gasses- $\begin{aligned} & \frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\text {total }}}=\mathrm{x}_{1} \\ & \mathrm{P}_{\mathrm{A}}=\mathrm{x}_{1} \times \mathrm{P}_{\text {total }} \quad \ldots \ldots . . \text {. (i) } \end{aligned}$ According to Raoult's law $\mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{x}_{2}$ From equation (i) and equation (ii) we get:- $\begin{aligned} & \mathrm{P}_{\text {total }} \mathrm{x}_{1}=\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{x}_{2} \\ & \mathrm{P}_{\text {total }}=\frac{\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{x}_{2}}{\mathrm{x}_{1}} \end{aligned}$
AMU-2018
SOLUTIONS
277330
A $0.0020 \mathrm{~m}$ aqueous solution of an ionic compound freezes at $-0.00732^{\circ} \mathrm{C}$. Number of moles of in which one mole of the ionic compound produces on being dissolved in water will be $\left(K_{f}=1.88 \mathrm{~K} \mathrm{kgm}^{-1}\right)$
277326
The latest heat of vaporization of a liquid at $227^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure is $12 \mathrm{kcal} \mathrm{mol}^{-1}$. The change in internal energy, if 3 moles of the liquid changes to vapour at same conditions is
277370
The aqueous solution that has the lowest vapour pressure at a given temperature is
1 0.1 molal sodium phosphate
2 0.1 molal barium chloride
3 0.1 molal sodium chloride
4 0.1 molal glucose
Explanation:
The same concentrations of relative lowering of vapour pressure depends upon the number of ions or particles present in different solutions. $\mathrm{Na}_{3} \mathrm{PO}_{4} \rightarrow 3 \mathrm{Na}^{+}+\mathrm{PO}_{4}^{3-}=4$ ions $\mathrm{BaCl}_{2} \rightarrow \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-}=3$ ions $\mathrm{NaCl} \rightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}=2$ ions Glucose $\rightarrow$ no dissociation $=1$ ions As $\mathrm{Na}_{3} \mathrm{PO}_{4}$ gives maximum ions (4) on dissociation, hence relative lowering of vapour pressure is maximum for sodium phosphate $\left(\mathrm{Na}_{3} \mathrm{PO}_{4}\right)$ and it has lowest vapour pressure.
UP CPMT-2013
SOLUTIONS
277329
If $x_{1}$ and $x_{2}$ represent the mole fraction of a component $A$ in the vapour phase and liquid mixture respectively and $p_{A}^{0}$ and $p_{B}^{0}$ represent Vapour pressure of pure $A$ and $B$, then total of vapour pressure of liquid mixture is
According to Dalton's law of partial pressures of gasses- $\begin{aligned} & \frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\text {total }}}=\mathrm{x}_{1} \\ & \mathrm{P}_{\mathrm{A}}=\mathrm{x}_{1} \times \mathrm{P}_{\text {total }} \quad \ldots \ldots . . \text {. (i) } \end{aligned}$ According to Raoult's law $\mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{x}_{2}$ From equation (i) and equation (ii) we get:- $\begin{aligned} & \mathrm{P}_{\text {total }} \mathrm{x}_{1}=\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{x}_{2} \\ & \mathrm{P}_{\text {total }}=\frac{\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{x}_{2}}{\mathrm{x}_{1}} \end{aligned}$
AMU-2018
SOLUTIONS
277330
A $0.0020 \mathrm{~m}$ aqueous solution of an ionic compound freezes at $-0.00732^{\circ} \mathrm{C}$. Number of moles of in which one mole of the ionic compound produces on being dissolved in water will be $\left(K_{f}=1.88 \mathrm{~K} \mathrm{kgm}^{-1}\right)$
277326
The latest heat of vaporization of a liquid at $227^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure is $12 \mathrm{kcal} \mathrm{mol}^{-1}$. The change in internal energy, if 3 moles of the liquid changes to vapour at same conditions is
277370
The aqueous solution that has the lowest vapour pressure at a given temperature is
1 0.1 molal sodium phosphate
2 0.1 molal barium chloride
3 0.1 molal sodium chloride
4 0.1 molal glucose
Explanation:
The same concentrations of relative lowering of vapour pressure depends upon the number of ions or particles present in different solutions. $\mathrm{Na}_{3} \mathrm{PO}_{4} \rightarrow 3 \mathrm{Na}^{+}+\mathrm{PO}_{4}^{3-}=4$ ions $\mathrm{BaCl}_{2} \rightarrow \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-}=3$ ions $\mathrm{NaCl} \rightarrow \mathrm{Na}^{+}+\mathrm{Cl}^{-}=2$ ions Glucose $\rightarrow$ no dissociation $=1$ ions As $\mathrm{Na}_{3} \mathrm{PO}_{4}$ gives maximum ions (4) on dissociation, hence relative lowering of vapour pressure is maximum for sodium phosphate $\left(\mathrm{Na}_{3} \mathrm{PO}_{4}\right)$ and it has lowest vapour pressure.
UP CPMT-2013
SOLUTIONS
277329
If $x_{1}$ and $x_{2}$ represent the mole fraction of a component $A$ in the vapour phase and liquid mixture respectively and $p_{A}^{0}$ and $p_{B}^{0}$ represent Vapour pressure of pure $A$ and $B$, then total of vapour pressure of liquid mixture is
According to Dalton's law of partial pressures of gasses- $\begin{aligned} & \frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\text {total }}}=\mathrm{x}_{1} \\ & \mathrm{P}_{\mathrm{A}}=\mathrm{x}_{1} \times \mathrm{P}_{\text {total }} \quad \ldots \ldots . . \text {. (i) } \end{aligned}$ According to Raoult's law $\mathrm{P}_{\mathrm{A}}=\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{x}_{2}$ From equation (i) and equation (ii) we get:- $\begin{aligned} & \mathrm{P}_{\text {total }} \mathrm{x}_{1}=\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{x}_{2} \\ & \mathrm{P}_{\text {total }}=\frac{\mathrm{P}_{\mathrm{A}}^{\circ} \mathrm{x}_{2}}{\mathrm{x}_{1}} \end{aligned}$
AMU-2018
SOLUTIONS
277330
A $0.0020 \mathrm{~m}$ aqueous solution of an ionic compound freezes at $-0.00732^{\circ} \mathrm{C}$. Number of moles of in which one mole of the ionic compound produces on being dissolved in water will be $\left(K_{f}=1.88 \mathrm{~K} \mathrm{kgm}^{-1}\right)$
