277275
0.01 $\mathrm{M}$ solution of $\mathrm{KCl}$ and $\mathrm{BaCl}_{2}$ are prepared in water. The freezing point of $\mathrm{KCl}$ is found to be $-2^{\circ} \mathrm{C}$. What is the freezing point of $\mathrm{BaCl}_{2}$ to be completely ionised?
1 $-3^{\circ} \mathrm{C}$
2 $+3^{\circ} \mathrm{C}$
3 $-2^{\circ} \mathrm{C}$
4 $-4^{\circ} \mathrm{C}$
Explanation:
Ionic reaction, $\begin{aligned} & \mathrm{KCl} \longrightarrow \mathrm{K}^{+}+\mathrm{Cl}^{-} \quad(\text { Here, } \mathrm{i}=2) \\ & \mathrm{BaCl}_{2} \longrightarrow \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-} \quad(\text { Here }, \mathrm{i}=3) \end{aligned}$ As, we know that, $\Delta \mathrm{T}_{\mathrm{f}} \propto \mathrm{i}$ $\Delta \mathrm{T}_{\mathrm{f}}$ is directly proportional to the molality of the solution i $\Delta \mathrm{T}_{\mathrm{f}}(\mathrm{KCl})=-2^{\circ} \mathrm{C}$ Now, $\begin{gathered} \frac{\Delta \mathrm{T}_{\mathrm{f}}(\mathrm{KCl})}{\Delta \mathrm{T}_{\mathrm{f}}\left(\mathrm{BaCl}_{2}\right)}=\frac{\mathrm{i}(\mathrm{KCl})}{\mathrm{i}\left(\mathrm{BaCl}_{2}\right)} \\ \Rightarrow \frac{\Delta \mathrm{T}_{\mathrm{f}}(\mathrm{KCl})}{\Delta \mathrm{T}_{\mathrm{f}}\left(\mathrm{BaCl}_{2}\right)}=\frac{2}{3} \\ \Rightarrow \Delta \mathrm{T}_{\mathrm{f}}\left(\mathrm{BaCl}_{2}\right)=\frac{3 \times(-2)}{2} \\ =-3^{\circ} \mathrm{C} \end{gathered}$
AIIMS-2008
SOLUTIONS
277276
The change in entropy for the fusion of 1 mole of ice is [m.p. of ice $=273 \mathrm{~K}$, molar enthalpy of fusion for ice $=\mathbf{6 . 0} \mathrm{kJ} \mathrm{mol}^{-1}$ ]
276965
If active mass of a $6 \%$ solution of a compound is 2 , its molecular weight will be
1 30
2 15
3 60
4 22
Explanation:
Given, The percentage of solute in the solution $=6 \%$ Mass of the compound $=2$ moles $/$ liter As $100 \mathrm{ml}$ solution contains $6 \mathrm{gm}$ of solute, $\therefore 1000 \mathrm{ml}$ Solution contains $=\frac{6 \times 1000}{100}=60 \mathrm{gm}$ of solute Thus, molecular weight of the compound $=\frac{60}{2}=30$
COMEDK-2020
SOLUTIONS
276968
$10 \mathrm{~g}$ of $\mathrm{NaOH}$ is dissolved in $500 \mathrm{~mL}$ of aqueous solution. Calculate the molarity of this solution ? (Given, formula weight of $\mathrm{NaOH}=40$ )
277275
0.01 $\mathrm{M}$ solution of $\mathrm{KCl}$ and $\mathrm{BaCl}_{2}$ are prepared in water. The freezing point of $\mathrm{KCl}$ is found to be $-2^{\circ} \mathrm{C}$. What is the freezing point of $\mathrm{BaCl}_{2}$ to be completely ionised?
1 $-3^{\circ} \mathrm{C}$
2 $+3^{\circ} \mathrm{C}$
3 $-2^{\circ} \mathrm{C}$
4 $-4^{\circ} \mathrm{C}$
Explanation:
Ionic reaction, $\begin{aligned} & \mathrm{KCl} \longrightarrow \mathrm{K}^{+}+\mathrm{Cl}^{-} \quad(\text { Here, } \mathrm{i}=2) \\ & \mathrm{BaCl}_{2} \longrightarrow \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-} \quad(\text { Here }, \mathrm{i}=3) \end{aligned}$ As, we know that, $\Delta \mathrm{T}_{\mathrm{f}} \propto \mathrm{i}$ $\Delta \mathrm{T}_{\mathrm{f}}$ is directly proportional to the molality of the solution i $\Delta \mathrm{T}_{\mathrm{f}}(\mathrm{KCl})=-2^{\circ} \mathrm{C}$ Now, $\begin{gathered} \frac{\Delta \mathrm{T}_{\mathrm{f}}(\mathrm{KCl})}{\Delta \mathrm{T}_{\mathrm{f}}\left(\mathrm{BaCl}_{2}\right)}=\frac{\mathrm{i}(\mathrm{KCl})}{\mathrm{i}\left(\mathrm{BaCl}_{2}\right)} \\ \Rightarrow \frac{\Delta \mathrm{T}_{\mathrm{f}}(\mathrm{KCl})}{\Delta \mathrm{T}_{\mathrm{f}}\left(\mathrm{BaCl}_{2}\right)}=\frac{2}{3} \\ \Rightarrow \Delta \mathrm{T}_{\mathrm{f}}\left(\mathrm{BaCl}_{2}\right)=\frac{3 \times(-2)}{2} \\ =-3^{\circ} \mathrm{C} \end{gathered}$
AIIMS-2008
SOLUTIONS
277276
The change in entropy for the fusion of 1 mole of ice is [m.p. of ice $=273 \mathrm{~K}$, molar enthalpy of fusion for ice $=\mathbf{6 . 0} \mathrm{kJ} \mathrm{mol}^{-1}$ ]
276965
If active mass of a $6 \%$ solution of a compound is 2 , its molecular weight will be
1 30
2 15
3 60
4 22
Explanation:
Given, The percentage of solute in the solution $=6 \%$ Mass of the compound $=2$ moles $/$ liter As $100 \mathrm{ml}$ solution contains $6 \mathrm{gm}$ of solute, $\therefore 1000 \mathrm{ml}$ Solution contains $=\frac{6 \times 1000}{100}=60 \mathrm{gm}$ of solute Thus, molecular weight of the compound $=\frac{60}{2}=30$
COMEDK-2020
SOLUTIONS
276968
$10 \mathrm{~g}$ of $\mathrm{NaOH}$ is dissolved in $500 \mathrm{~mL}$ of aqueous solution. Calculate the molarity of this solution ? (Given, formula weight of $\mathrm{NaOH}=40$ )
277275
0.01 $\mathrm{M}$ solution of $\mathrm{KCl}$ and $\mathrm{BaCl}_{2}$ are prepared in water. The freezing point of $\mathrm{KCl}$ is found to be $-2^{\circ} \mathrm{C}$. What is the freezing point of $\mathrm{BaCl}_{2}$ to be completely ionised?
1 $-3^{\circ} \mathrm{C}$
2 $+3^{\circ} \mathrm{C}$
3 $-2^{\circ} \mathrm{C}$
4 $-4^{\circ} \mathrm{C}$
Explanation:
Ionic reaction, $\begin{aligned} & \mathrm{KCl} \longrightarrow \mathrm{K}^{+}+\mathrm{Cl}^{-} \quad(\text { Here, } \mathrm{i}=2) \\ & \mathrm{BaCl}_{2} \longrightarrow \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-} \quad(\text { Here }, \mathrm{i}=3) \end{aligned}$ As, we know that, $\Delta \mathrm{T}_{\mathrm{f}} \propto \mathrm{i}$ $\Delta \mathrm{T}_{\mathrm{f}}$ is directly proportional to the molality of the solution i $\Delta \mathrm{T}_{\mathrm{f}}(\mathrm{KCl})=-2^{\circ} \mathrm{C}$ Now, $\begin{gathered} \frac{\Delta \mathrm{T}_{\mathrm{f}}(\mathrm{KCl})}{\Delta \mathrm{T}_{\mathrm{f}}\left(\mathrm{BaCl}_{2}\right)}=\frac{\mathrm{i}(\mathrm{KCl})}{\mathrm{i}\left(\mathrm{BaCl}_{2}\right)} \\ \Rightarrow \frac{\Delta \mathrm{T}_{\mathrm{f}}(\mathrm{KCl})}{\Delta \mathrm{T}_{\mathrm{f}}\left(\mathrm{BaCl}_{2}\right)}=\frac{2}{3} \\ \Rightarrow \Delta \mathrm{T}_{\mathrm{f}}\left(\mathrm{BaCl}_{2}\right)=\frac{3 \times(-2)}{2} \\ =-3^{\circ} \mathrm{C} \end{gathered}$
AIIMS-2008
SOLUTIONS
277276
The change in entropy for the fusion of 1 mole of ice is [m.p. of ice $=273 \mathrm{~K}$, molar enthalpy of fusion for ice $=\mathbf{6 . 0} \mathrm{kJ} \mathrm{mol}^{-1}$ ]
276965
If active mass of a $6 \%$ solution of a compound is 2 , its molecular weight will be
1 30
2 15
3 60
4 22
Explanation:
Given, The percentage of solute in the solution $=6 \%$ Mass of the compound $=2$ moles $/$ liter As $100 \mathrm{ml}$ solution contains $6 \mathrm{gm}$ of solute, $\therefore 1000 \mathrm{ml}$ Solution contains $=\frac{6 \times 1000}{100}=60 \mathrm{gm}$ of solute Thus, molecular weight of the compound $=\frac{60}{2}=30$
COMEDK-2020
SOLUTIONS
276968
$10 \mathrm{~g}$ of $\mathrm{NaOH}$ is dissolved in $500 \mathrm{~mL}$ of aqueous solution. Calculate the molarity of this solution ? (Given, formula weight of $\mathrm{NaOH}=40$ )
277275
0.01 $\mathrm{M}$ solution of $\mathrm{KCl}$ and $\mathrm{BaCl}_{2}$ are prepared in water. The freezing point of $\mathrm{KCl}$ is found to be $-2^{\circ} \mathrm{C}$. What is the freezing point of $\mathrm{BaCl}_{2}$ to be completely ionised?
1 $-3^{\circ} \mathrm{C}$
2 $+3^{\circ} \mathrm{C}$
3 $-2^{\circ} \mathrm{C}$
4 $-4^{\circ} \mathrm{C}$
Explanation:
Ionic reaction, $\begin{aligned} & \mathrm{KCl} \longrightarrow \mathrm{K}^{+}+\mathrm{Cl}^{-} \quad(\text { Here, } \mathrm{i}=2) \\ & \mathrm{BaCl}_{2} \longrightarrow \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-} \quad(\text { Here }, \mathrm{i}=3) \end{aligned}$ As, we know that, $\Delta \mathrm{T}_{\mathrm{f}} \propto \mathrm{i}$ $\Delta \mathrm{T}_{\mathrm{f}}$ is directly proportional to the molality of the solution i $\Delta \mathrm{T}_{\mathrm{f}}(\mathrm{KCl})=-2^{\circ} \mathrm{C}$ Now, $\begin{gathered} \frac{\Delta \mathrm{T}_{\mathrm{f}}(\mathrm{KCl})}{\Delta \mathrm{T}_{\mathrm{f}}\left(\mathrm{BaCl}_{2}\right)}=\frac{\mathrm{i}(\mathrm{KCl})}{\mathrm{i}\left(\mathrm{BaCl}_{2}\right)} \\ \Rightarrow \frac{\Delta \mathrm{T}_{\mathrm{f}}(\mathrm{KCl})}{\Delta \mathrm{T}_{\mathrm{f}}\left(\mathrm{BaCl}_{2}\right)}=\frac{2}{3} \\ \Rightarrow \Delta \mathrm{T}_{\mathrm{f}}\left(\mathrm{BaCl}_{2}\right)=\frac{3 \times(-2)}{2} \\ =-3^{\circ} \mathrm{C} \end{gathered}$
AIIMS-2008
SOLUTIONS
277276
The change in entropy for the fusion of 1 mole of ice is [m.p. of ice $=273 \mathrm{~K}$, molar enthalpy of fusion for ice $=\mathbf{6 . 0} \mathrm{kJ} \mathrm{mol}^{-1}$ ]
276965
If active mass of a $6 \%$ solution of a compound is 2 , its molecular weight will be
1 30
2 15
3 60
4 22
Explanation:
Given, The percentage of solute in the solution $=6 \%$ Mass of the compound $=2$ moles $/$ liter As $100 \mathrm{ml}$ solution contains $6 \mathrm{gm}$ of solute, $\therefore 1000 \mathrm{ml}$ Solution contains $=\frac{6 \times 1000}{100}=60 \mathrm{gm}$ of solute Thus, molecular weight of the compound $=\frac{60}{2}=30$
COMEDK-2020
SOLUTIONS
276968
$10 \mathrm{~g}$ of $\mathrm{NaOH}$ is dissolved in $500 \mathrm{~mL}$ of aqueous solution. Calculate the molarity of this solution ? (Given, formula weight of $\mathrm{NaOH}=40$ )
277275
0.01 $\mathrm{M}$ solution of $\mathrm{KCl}$ and $\mathrm{BaCl}_{2}$ are prepared in water. The freezing point of $\mathrm{KCl}$ is found to be $-2^{\circ} \mathrm{C}$. What is the freezing point of $\mathrm{BaCl}_{2}$ to be completely ionised?
1 $-3^{\circ} \mathrm{C}$
2 $+3^{\circ} \mathrm{C}$
3 $-2^{\circ} \mathrm{C}$
4 $-4^{\circ} \mathrm{C}$
Explanation:
Ionic reaction, $\begin{aligned} & \mathrm{KCl} \longrightarrow \mathrm{K}^{+}+\mathrm{Cl}^{-} \quad(\text { Here, } \mathrm{i}=2) \\ & \mathrm{BaCl}_{2} \longrightarrow \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-} \quad(\text { Here }, \mathrm{i}=3) \end{aligned}$ As, we know that, $\Delta \mathrm{T}_{\mathrm{f}} \propto \mathrm{i}$ $\Delta \mathrm{T}_{\mathrm{f}}$ is directly proportional to the molality of the solution i $\Delta \mathrm{T}_{\mathrm{f}}(\mathrm{KCl})=-2^{\circ} \mathrm{C}$ Now, $\begin{gathered} \frac{\Delta \mathrm{T}_{\mathrm{f}}(\mathrm{KCl})}{\Delta \mathrm{T}_{\mathrm{f}}\left(\mathrm{BaCl}_{2}\right)}=\frac{\mathrm{i}(\mathrm{KCl})}{\mathrm{i}\left(\mathrm{BaCl}_{2}\right)} \\ \Rightarrow \frac{\Delta \mathrm{T}_{\mathrm{f}}(\mathrm{KCl})}{\Delta \mathrm{T}_{\mathrm{f}}\left(\mathrm{BaCl}_{2}\right)}=\frac{2}{3} \\ \Rightarrow \Delta \mathrm{T}_{\mathrm{f}}\left(\mathrm{BaCl}_{2}\right)=\frac{3 \times(-2)}{2} \\ =-3^{\circ} \mathrm{C} \end{gathered}$
AIIMS-2008
SOLUTIONS
277276
The change in entropy for the fusion of 1 mole of ice is [m.p. of ice $=273 \mathrm{~K}$, molar enthalpy of fusion for ice $=\mathbf{6 . 0} \mathrm{kJ} \mathrm{mol}^{-1}$ ]
276965
If active mass of a $6 \%$ solution of a compound is 2 , its molecular weight will be
1 30
2 15
3 60
4 22
Explanation:
Given, The percentage of solute in the solution $=6 \%$ Mass of the compound $=2$ moles $/$ liter As $100 \mathrm{ml}$ solution contains $6 \mathrm{gm}$ of solute, $\therefore 1000 \mathrm{ml}$ Solution contains $=\frac{6 \times 1000}{100}=60 \mathrm{gm}$ of solute Thus, molecular weight of the compound $=\frac{60}{2}=30$
COMEDK-2020
SOLUTIONS
276968
$10 \mathrm{~g}$ of $\mathrm{NaOH}$ is dissolved in $500 \mathrm{~mL}$ of aqueous solution. Calculate the molarity of this solution ? (Given, formula weight of $\mathrm{NaOH}=40$ )
