QBManager

SOLUTIONS

277275 0.01 $M$ solution of $KCl$ and ${BaCl}_{2}$ are prepared in water. The freezing point of $KCl$ is found to be $- 2^{\circ} C$ . What is the freezing point of ${BaCl}_{2}$ to be completely ionised?

1

- 3^{\circ} C

2

+ 3^{\circ} C

3

- 2^{\circ} C

4

- 4^{\circ} C

Explanation:

Ionic reaction,
$\begin{aligned} KCl ⟶ K^{+} + {Cl}^{-} (Here, i = 2) \\ {BaCl}_{2} ⟶ {Ba}^{2 +} + 2 {Cl}^{-} (Here, i = 3) \end{aligned}$
As, we know that,
$Δ T_{f} \propto i$
$Δ T_{f}$ is directly proportional to the molality of the solution i
$Δ T_{f} (KCl) = - 2^{\circ} C$
Now,
$\begin{matrix} \frac{Δ T_{f} (KCl)}{Δ T_{f} ({BaCl}_{2})} = \frac{i (KCl)}{i ({BaCl}_{2})} \\ \Rightarrow \frac{Δ T_{f} (KCl)}{Δ T_{f} ({BaCl}_{2})} = \frac{2}{3} \\ \Rightarrow Δ T_{f} ({BaCl}_{2}) = \frac{3 \times (- 2)}{2} \\ = - 3^{\circ} C \end{matrix}$

AIIMS-2008

SOLUTIONS

277276 The change in entropy for the fusion of 1 mole of ice is [m.p. of ice $= 273 K$ , molar enthalpy of fusion for ice $= 6.0 kJ {mol}^{- 1}$ ]

1

11.73 {JK}^{- 1} {mol}^{- 1}

2

18.84 {JK}^{- 1} {mol}^{- 1}

3

21.97 {JK}^{- 1} {mol}^{- 1}

4

24.47 {JK}^{- 1} {mol}^{- 1}

Explanation:

Given data,
Molar enthalpy of fusion for ice $= 6.0 kJ {mol}^{- 1}$
$Δ H_{fusion.} = 6.0 \times 10^{3} J {mol}^{- 1}$
$∴ S_{fusion} = \frac{Δ H_{fusion}}{T_{m.p.}}$
$= \frac{6.0 \times 10^{3} {Jmol}^{- 1}}{273 K}$
$= 21.976 {JK}^{- 1} {mol}^{- 1}$

JCECE-2007

SOLUTIONS

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 $100 ml$ solution contains $6 gm$ of solute,
$∴ 1000 ml$ Solution contains $= \frac{6 \times 1000}{100} = 60 gm$ of solute
Thus, molecular weight of the compound $= \frac{60}{2} = 30$

COMEDK-2020

SOLUTIONS

276968 $10 g$ of $NaOH$ is dissolved in $500 mL$ of aqueous solution. Calculate the molarity of this solution ? (Given, formula weight of $NaOH = 40$ )

1

0.5 \times 10^{- 3} M

2

0.4 M

3

0.25 \times 10^{- 3} M

4

0.5 M

Explanation:

Molarity $= \frac{Moles of solute}{Volume of solution (in litre)}$
$\begin{aligned} = \frac{Weight of solute \times 1000}{Molecular weight solute \times Volume (mL)} \\ = \frac{10 \times 1000}{40 \times 500} = 0.5 M \end{aligned}$

AP EAMCET (Engg.) 17.09.2020 Shift-I

SOLUTIONS

276969 A $40 % HCl$ solution has density $1.2 g {mL}^{- 1}$ . The molarity of the solution is nearly

1

11 M

2

12 M

3

13 M

4

14 M

Explanation:

As, density (d) of $HC ℓ$ solution is given (pc) $(1.2 g {mL}^{- 1})$ , the percentage strength of the solution $(40 %)$ is of $(w / w)$ type.
$\begin{aligned} \Rightarrow Molarity & = \frac{pc (w / w) \times d ({gmL}^{- 1}) \times 10}{M_{HC ℓ}} \\ = \frac{40 \times 1.2 \times 10}{36.5} = 13.15 ≃ 13 M \end{aligned}$

AP EAMCET (Engg.) 17.09.2020 Shift-I

SOLUTIONS

277275 0.01 $M$ solution of $KCl$ and ${BaCl}_{2}$ are prepared in water. The freezing point of $KCl$ is found to be $- 2^{\circ} C$ . What is the freezing point of ${BaCl}_{2}$ to be completely ionised?

1

- 3^{\circ} C

2

+ 3^{\circ} C

3

- 2^{\circ} C

4

- 4^{\circ} C

Explanation:

Ionic reaction,
$\begin{aligned} KCl ⟶ K^{+} + {Cl}^{-} (Here, i = 2) \\ {BaCl}_{2} ⟶ {Ba}^{2 +} + 2 {Cl}^{-} (Here, i = 3) \end{aligned}$
As, we know that,
$Δ T_{f} \propto i$
$Δ T_{f}$ is directly proportional to the molality of the solution i
$Δ T_{f} (KCl) = - 2^{\circ} C$
Now,
$\begin{matrix} \frac{Δ T_{f} (KCl)}{Δ T_{f} ({BaCl}_{2})} = \frac{i (KCl)}{i ({BaCl}_{2})} \\ \Rightarrow \frac{Δ T_{f} (KCl)}{Δ T_{f} ({BaCl}_{2})} = \frac{2}{3} \\ \Rightarrow Δ T_{f} ({BaCl}_{2}) = \frac{3 \times (- 2)}{2} \\ = - 3^{\circ} C \end{matrix}$

AIIMS-2008

SOLUTIONS

277276 The change in entropy for the fusion of 1 mole of ice is [m.p. of ice $= 273 K$ , molar enthalpy of fusion for ice $= 6.0 kJ {mol}^{- 1}$ ]

1

11.73 {JK}^{- 1} {mol}^{- 1}

2

18.84 {JK}^{- 1} {mol}^{- 1}

3

21.97 {JK}^{- 1} {mol}^{- 1}

4

24.47 {JK}^{- 1} {mol}^{- 1}

Explanation:

Given data,
Molar enthalpy of fusion for ice $= 6.0 kJ {mol}^{- 1}$
$Δ H_{fusion.} = 6.0 \times 10^{3} J {mol}^{- 1}$
$∴ S_{fusion} = \frac{Δ H_{fusion}}{T_{m.p.}}$
$= \frac{6.0 \times 10^{3} {Jmol}^{- 1}}{273 K}$
$= 21.976 {JK}^{- 1} {mol}^{- 1}$

JCECE-2007

SOLUTIONS

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 $100 ml$ solution contains $6 gm$ of solute,
$∴ 1000 ml$ Solution contains $= \frac{6 \times 1000}{100} = 60 gm$ of solute
Thus, molecular weight of the compound $= \frac{60}{2} = 30$

COMEDK-2020

SOLUTIONS

276968 $10 g$ of $NaOH$ is dissolved in $500 mL$ of aqueous solution. Calculate the molarity of this solution ? (Given, formula weight of $NaOH = 40$ )

1

0.5 \times 10^{- 3} M

2

0.4 M

3

0.25 \times 10^{- 3} M

4

0.5 M

Explanation:

Molarity $= \frac{Moles of solute}{Volume of solution (in litre)}$
$\begin{aligned} = \frac{Weight of solute \times 1000}{Molecular weight solute \times Volume (mL)} \\ = \frac{10 \times 1000}{40 \times 500} = 0.5 M \end{aligned}$

AP EAMCET (Engg.) 17.09.2020 Shift-I

SOLUTIONS

276969 A $40 % HCl$ solution has density $1.2 g {mL}^{- 1}$ . The molarity of the solution is nearly

1

11 M

2

12 M

3

13 M

4

14 M

Explanation:

As, density (d) of $HC ℓ$ solution is given (pc) $(1.2 g {mL}^{- 1})$ , the percentage strength of the solution $(40 %)$ is of $(w / w)$ type.
$\begin{aligned} \Rightarrow Molarity & = \frac{pc (w / w) \times d ({gmL}^{- 1}) \times 10}{M_{HC ℓ}} \\ = \frac{40 \times 1.2 \times 10}{36.5} = 13.15 ≃ 13 M \end{aligned}$

AP EAMCET (Engg.) 17.09.2020 Shift-I

SOLUTIONS

277275 0.01 $M$ solution of $KCl$ and ${BaCl}_{2}$ are prepared in water. The freezing point of $KCl$ is found to be $- 2^{\circ} C$ . What is the freezing point of ${BaCl}_{2}$ to be completely ionised?

1

- 3^{\circ} C

2

+ 3^{\circ} C

3

- 2^{\circ} C

4

- 4^{\circ} C

Explanation:

Ionic reaction,
$\begin{aligned} KCl ⟶ K^{+} + {Cl}^{-} (Here, i = 2) \\ {BaCl}_{2} ⟶ {Ba}^{2 +} + 2 {Cl}^{-} (Here, i = 3) \end{aligned}$
As, we know that,
$Δ T_{f} \propto i$
$Δ T_{f}$ is directly proportional to the molality of the solution i
$Δ T_{f} (KCl) = - 2^{\circ} C$
Now,
$\begin{matrix} \frac{Δ T_{f} (KCl)}{Δ T_{f} ({BaCl}_{2})} = \frac{i (KCl)}{i ({BaCl}_{2})} \\ \Rightarrow \frac{Δ T_{f} (KCl)}{Δ T_{f} ({BaCl}_{2})} = \frac{2}{3} \\ \Rightarrow Δ T_{f} ({BaCl}_{2}) = \frac{3 \times (- 2)}{2} \\ = - 3^{\circ} C \end{matrix}$

AIIMS-2008

SOLUTIONS

277276 The change in entropy for the fusion of 1 mole of ice is [m.p. of ice $= 273 K$ , molar enthalpy of fusion for ice $= 6.0 kJ {mol}^{- 1}$ ]

1

11.73 {JK}^{- 1} {mol}^{- 1}

2

18.84 {JK}^{- 1} {mol}^{- 1}

3

21.97 {JK}^{- 1} {mol}^{- 1}

4

24.47 {JK}^{- 1} {mol}^{- 1}

Explanation:

Given data,
Molar enthalpy of fusion for ice $= 6.0 kJ {mol}^{- 1}$
$Δ H_{fusion.} = 6.0 \times 10^{3} J {mol}^{- 1}$
$∴ S_{fusion} = \frac{Δ H_{fusion}}{T_{m.p.}}$
$= \frac{6.0 \times 10^{3} {Jmol}^{- 1}}{273 K}$
$= 21.976 {JK}^{- 1} {mol}^{- 1}$

JCECE-2007

SOLUTIONS

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 $100 ml$ solution contains $6 gm$ of solute,
$∴ 1000 ml$ Solution contains $= \frac{6 \times 1000}{100} = 60 gm$ of solute
Thus, molecular weight of the compound $= \frac{60}{2} = 30$

COMEDK-2020

SOLUTIONS

276968 $10 g$ of $NaOH$ is dissolved in $500 mL$ of aqueous solution. Calculate the molarity of this solution ? (Given, formula weight of $NaOH = 40$ )

1

0.5 \times 10^{- 3} M

2

0.4 M

3

0.25 \times 10^{- 3} M

4

0.5 M

Explanation:

Molarity $= \frac{Moles of solute}{Volume of solution (in litre)}$
$\begin{aligned} = \frac{Weight of solute \times 1000}{Molecular weight solute \times Volume (mL)} \\ = \frac{10 \times 1000}{40 \times 500} = 0.5 M \end{aligned}$

AP EAMCET (Engg.) 17.09.2020 Shift-I

SOLUTIONS

276969 A $40 % HCl$ solution has density $1.2 g {mL}^{- 1}$ . The molarity of the solution is nearly

1

11 M

2

12 M

3

13 M

4

14 M

Explanation:

As, density (d) of $HC ℓ$ solution is given (pc) $(1.2 g {mL}^{- 1})$ , the percentage strength of the solution $(40 %)$ is of $(w / w)$ type.
$\begin{aligned} \Rightarrow Molarity & = \frac{pc (w / w) \times d ({gmL}^{- 1}) \times 10}{M_{HC ℓ}} \\ = \frac{40 \times 1.2 \times 10}{36.5} = 13.15 ≃ 13 M \end{aligned}$

AP EAMCET (Engg.) 17.09.2020 Shift-I

SOLUTIONS

277275 0.01 $M$ solution of $KCl$ and ${BaCl}_{2}$ are prepared in water. The freezing point of $KCl$ is found to be $- 2^{\circ} C$ . What is the freezing point of ${BaCl}_{2}$ to be completely ionised?

1

- 3^{\circ} C

2

+ 3^{\circ} C

3

- 2^{\circ} C

4

- 4^{\circ} C

Explanation:

Ionic reaction,
$\begin{aligned} KCl ⟶ K^{+} + {Cl}^{-} (Here, i = 2) \\ {BaCl}_{2} ⟶ {Ba}^{2 +} + 2 {Cl}^{-} (Here, i = 3) \end{aligned}$
As, we know that,
$Δ T_{f} \propto i$
$Δ T_{f}$ is directly proportional to the molality of the solution i
$Δ T_{f} (KCl) = - 2^{\circ} C$
Now,
$\begin{matrix} \frac{Δ T_{f} (KCl)}{Δ T_{f} ({BaCl}_{2})} = \frac{i (KCl)}{i ({BaCl}_{2})} \\ \Rightarrow \frac{Δ T_{f} (KCl)}{Δ T_{f} ({BaCl}_{2})} = \frac{2}{3} \\ \Rightarrow Δ T_{f} ({BaCl}_{2}) = \frac{3 \times (- 2)}{2} \\ = - 3^{\circ} C \end{matrix}$

AIIMS-2008

SOLUTIONS

277276 The change in entropy for the fusion of 1 mole of ice is [m.p. of ice $= 273 K$ , molar enthalpy of fusion for ice $= 6.0 kJ {mol}^{- 1}$ ]

1

11.73 {JK}^{- 1} {mol}^{- 1}

2

18.84 {JK}^{- 1} {mol}^{- 1}

3

21.97 {JK}^{- 1} {mol}^{- 1}

4

24.47 {JK}^{- 1} {mol}^{- 1}

Explanation:

Given data,
Molar enthalpy of fusion for ice $= 6.0 kJ {mol}^{- 1}$
$Δ H_{fusion.} = 6.0 \times 10^{3} J {mol}^{- 1}$
$∴ S_{fusion} = \frac{Δ H_{fusion}}{T_{m.p.}}$
$= \frac{6.0 \times 10^{3} {Jmol}^{- 1}}{273 K}$
$= 21.976 {JK}^{- 1} {mol}^{- 1}$

JCECE-2007

SOLUTIONS

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 $100 ml$ solution contains $6 gm$ of solute,
$∴ 1000 ml$ Solution contains $= \frac{6 \times 1000}{100} = 60 gm$ of solute
Thus, molecular weight of the compound $= \frac{60}{2} = 30$

COMEDK-2020

SOLUTIONS

276968 $10 g$ of $NaOH$ is dissolved in $500 mL$ of aqueous solution. Calculate the molarity of this solution ? (Given, formula weight of $NaOH = 40$ )

1

0.5 \times 10^{- 3} M

2

0.4 M

3

0.25 \times 10^{- 3} M

4

0.5 M

Explanation:

Molarity $= \frac{Moles of solute}{Volume of solution (in litre)}$
$\begin{aligned} = \frac{Weight of solute \times 1000}{Molecular weight solute \times Volume (mL)} \\ = \frac{10 \times 1000}{40 \times 500} = 0.5 M \end{aligned}$

AP EAMCET (Engg.) 17.09.2020 Shift-I

SOLUTIONS

276969 A $40 % HCl$ solution has density $1.2 g {mL}^{- 1}$ . The molarity of the solution is nearly

1

11 M

2

12 M

3

13 M

4

14 M

Explanation:

As, density (d) of $HC ℓ$ solution is given (pc) $(1.2 g {mL}^{- 1})$ , the percentage strength of the solution $(40 %)$ is of $(w / w)$ type.
$\begin{aligned} \Rightarrow Molarity & = \frac{pc (w / w) \times d ({gmL}^{- 1}) \times 10}{M_{HC ℓ}} \\ = \frac{40 \times 1.2 \times 10}{36.5} = 13.15 ≃ 13 M \end{aligned}$

AP EAMCET (Engg.) 17.09.2020 Shift-I

SOLUTIONS

277275 0.01 $M$ solution of $KCl$ and ${BaCl}_{2}$ are prepared in water. The freezing point of $KCl$ is found to be $- 2^{\circ} C$ . What is the freezing point of ${BaCl}_{2}$ to be completely ionised?

1

- 3^{\circ} C

2

+ 3^{\circ} C

3

- 2^{\circ} C

4

- 4^{\circ} C

Explanation:

Ionic reaction,
$\begin{aligned} KCl ⟶ K^{+} + {Cl}^{-} (Here, i = 2) \\ {BaCl}_{2} ⟶ {Ba}^{2 +} + 2 {Cl}^{-} (Here, i = 3) \end{aligned}$
As, we know that,
$Δ T_{f} \propto i$
$Δ T_{f}$ is directly proportional to the molality of the solution i
$Δ T_{f} (KCl) = - 2^{\circ} C$
Now,
$\begin{matrix} \frac{Δ T_{f} (KCl)}{Δ T_{f} ({BaCl}_{2})} = \frac{i (KCl)}{i ({BaCl}_{2})} \\ \Rightarrow \frac{Δ T_{f} (KCl)}{Δ T_{f} ({BaCl}_{2})} = \frac{2}{3} \\ \Rightarrow Δ T_{f} ({BaCl}_{2}) = \frac{3 \times (- 2)}{2} \\ = - 3^{\circ} C \end{matrix}$

AIIMS-2008

SOLUTIONS

277276 The change in entropy for the fusion of 1 mole of ice is [m.p. of ice $= 273 K$ , molar enthalpy of fusion for ice $= 6.0 kJ {mol}^{- 1}$ ]

1

11.73 {JK}^{- 1} {mol}^{- 1}

2

18.84 {JK}^{- 1} {mol}^{- 1}

3

21.97 {JK}^{- 1} {mol}^{- 1}

4

24.47 {JK}^{- 1} {mol}^{- 1}

Explanation:

Given data,
Molar enthalpy of fusion for ice $= 6.0 kJ {mol}^{- 1}$
$Δ H_{fusion.} = 6.0 \times 10^{3} J {mol}^{- 1}$
$∴ S_{fusion} = \frac{Δ H_{fusion}}{T_{m.p.}}$
$= \frac{6.0 \times 10^{3} {Jmol}^{- 1}}{273 K}$
$= 21.976 {JK}^{- 1} {mol}^{- 1}$

JCECE-2007

SOLUTIONS

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 $100 ml$ solution contains $6 gm$ of solute,
$∴ 1000 ml$ Solution contains $= \frac{6 \times 1000}{100} = 60 gm$ of solute
Thus, molecular weight of the compound $= \frac{60}{2} = 30$

COMEDK-2020

SOLUTIONS

276968 $10 g$ of $NaOH$ is dissolved in $500 mL$ of aqueous solution. Calculate the molarity of this solution ? (Given, formula weight of $NaOH = 40$ )

1

0.5 \times 10^{- 3} M

2

0.4 M

3

0.25 \times 10^{- 3} M

4

0.5 M

Explanation:

Molarity $= \frac{Moles of solute}{Volume of solution (in litre)}$
$\begin{aligned} = \frac{Weight of solute \times 1000}{Molecular weight solute \times Volume (mL)} \\ = \frac{10 \times 1000}{40 \times 500} = 0.5 M \end{aligned}$

AP EAMCET (Engg.) 17.09.2020 Shift-I

SOLUTIONS

276969 A $40 % HCl$ solution has density $1.2 g {mL}^{- 1}$ . The molarity of the solution is nearly

1

11 M

2

12 M

3

13 M

4

14 M

Explanation:

As, density (d) of $HC ℓ$ solution is given (pc) $(1.2 g {mL}^{- 1})$ , the percentage strength of the solution $(40 %)$ is of $(w / w)$ type.
$\begin{aligned} \Rightarrow Molarity & = \frac{pc (w / w) \times d ({gmL}^{- 1}) \times 10}{M_{HC ℓ}} \\ = \frac{40 \times 1.2 \times 10}{36.5} = 13.15 ≃ 13 M \end{aligned}$

AP EAMCET (Engg.) 17.09.2020 Shift-I

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