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SOLUTIONS

277446 At $37^{\circ} C$ osmotic pressure of human blood is 7.65 atm. Tell how much glucose can be used in 1 lit of water for intravenous injection so that osmotic pressure of this glucose solution becomes equal to osmotic pressure of human blood.

1

22.2 gm

2

54.2 gm

3

15 gm

4

59.8 gm

Explanation:

Given that- $Π = 7.65$
$\begin{aligned} R = 0.08206 \\ T = 273 + 37 = 300 K \end{aligned}$
We know that-
$\begin{matrix} Π = CRT \\ 7.65 = C \times 0.08206 \times (273 + 37) \\ C = 0.3 M \end{matrix}$
Here, $1 L$ of solution will contain 0.3 moles of glucose
$= 0.3 \times 180 = 54 g$

Tripura JEE-2021

SOLUTIONS

277448 The osmotic pressure of a dilute solution of an ionic compound $XY$ in water is four times that of a solution of $0.01 M {BaCl}_{2}$ in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of $X Y$ (in mol $L^{- 1}$ ) in solution is

1

4 \times 10^{- 2}

2

16 \times 10^{- 4}

3

4 \times 10^{- 4}

4

6 \times 10^{- 2}

Explanation:

For xy
$i = 2 as complete dissociation takes place.$
$Π xy = 2 \times CRT$
For ${BaCl}_{2}$
$i = 3, Π {BaCl}_{2} = 3 \times 0.01 \times RT = 0.03 RT$
As $\prod xy = 4 \times \prod {BaCl}_{2}$
$\begin{aligned} \frac{\prod x y}{\prod {BaCl}_{2}} = \frac{2 \times CRT}{0.03 RT} \\ \frac{4 \prod {BaCl}_{2}}{\prod {BaCl}_{2}} = \frac{2 \times C}{0.03} \\ C = \frac{4 \times 0.03}{2} = 6 \times 10^{- 2} {molL}^{- 1} \end{aligned}$

[JEE Main 2019

SOLUTIONS

277449 Molal depression constant for a solvent is $4.0 K$ $kg mol^{- 1}$ . The depression in the freezing point of the solvent for $0.03 mol kg^{- 1}$ solution of $K_{2} {SO}_{4}$ is (Assume complete dissociation of the electrolyte)

1

0.18 K

2

0.36 K

3

0.12 K

4

0.24 K

Explanation:

Ionic Solution-
$K_{2} {SO}_{4} \to 2 K^{+} + {SO}^{2 -}$
Here $i = 3, m = 0.03 mol {kg}^{- 1}$
$Δ T_{f} = i \times k_{f} \times m$
$= 3 \times 4 \times 0.03$
$= 0.36 K$

[JEE Main 2019

SOLUTIONS

277450 A solution is prepared by dissolving $0.6 g$ of urea (molar mass $= 60 g {mol}^{- 1}$ ) and $1.8 g$ of glucose (molar mass $= 180 g {mol}^{- 1})$ in $100 mL$ of water at $27^{\circ} C$ . The osmotic pressure of the solution is $(R = 0.08206 L atm K {mol}^{- 1})$

1

8.2 atm

2

2.46 atm

3

4.92 atm

4

1.64 atm

Explanation:

From formula,
$Π_{1} = C_{1} RT$
$Π_{2} = C_{2} RT$
Hence,
$\begin{aligned} Π = Π_{1} + Π_{2} \\ = & \frac{0.6 / 60}{0.1} RT + \frac{1.8 / 180}{0.1} RT \\ Π & = 2 (0.1) RT \\ = 0.2 \times 24.6 \\ = & 4.92 atm . \end{aligned}$

[JEE Main 2019

SOLUTIONS

277446 At $37^{\circ} C$ osmotic pressure of human blood is 7.65 atm. Tell how much glucose can be used in 1 lit of water for intravenous injection so that osmotic pressure of this glucose solution becomes equal to osmotic pressure of human blood.

1

22.2 gm

2

54.2 gm

3

15 gm

4

59.8 gm

Explanation:

Given that- $Π = 7.65$
$\begin{aligned} R = 0.08206 \\ T = 273 + 37 = 300 K \end{aligned}$
We know that-
$\begin{matrix} Π = CRT \\ 7.65 = C \times 0.08206 \times (273 + 37) \\ C = 0.3 M \end{matrix}$
Here, $1 L$ of solution will contain 0.3 moles of glucose
$= 0.3 \times 180 = 54 g$

Tripura JEE-2021

SOLUTIONS

277448 The osmotic pressure of a dilute solution of an ionic compound $XY$ in water is four times that of a solution of $0.01 M {BaCl}_{2}$ in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of $X Y$ (in mol $L^{- 1}$ ) in solution is

1

4 \times 10^{- 2}

2

16 \times 10^{- 4}

3

4 \times 10^{- 4}

4

6 \times 10^{- 2}

Explanation:

For xy
$i = 2 as complete dissociation takes place.$
$Π xy = 2 \times CRT$
For ${BaCl}_{2}$
$i = 3, Π {BaCl}_{2} = 3 \times 0.01 \times RT = 0.03 RT$
As $\prod xy = 4 \times \prod {BaCl}_{2}$
$\begin{aligned} \frac{\prod x y}{\prod {BaCl}_{2}} = \frac{2 \times CRT}{0.03 RT} \\ \frac{4 \prod {BaCl}_{2}}{\prod {BaCl}_{2}} = \frac{2 \times C}{0.03} \\ C = \frac{4 \times 0.03}{2} = 6 \times 10^{- 2} {molL}^{- 1} \end{aligned}$

[JEE Main 2019

SOLUTIONS

277449 Molal depression constant for a solvent is $4.0 K$ $kg mol^{- 1}$ . The depression in the freezing point of the solvent for $0.03 mol kg^{- 1}$ solution of $K_{2} {SO}_{4}$ is (Assume complete dissociation of the electrolyte)

1

0.18 K

2

0.36 K

3

0.12 K

4

0.24 K

Explanation:

Ionic Solution-
$K_{2} {SO}_{4} \to 2 K^{+} + {SO}^{2 -}$
Here $i = 3, m = 0.03 mol {kg}^{- 1}$
$Δ T_{f} = i \times k_{f} \times m$
$= 3 \times 4 \times 0.03$
$= 0.36 K$

[JEE Main 2019

SOLUTIONS

277450 A solution is prepared by dissolving $0.6 g$ of urea (molar mass $= 60 g {mol}^{- 1}$ ) and $1.8 g$ of glucose (molar mass $= 180 g {mol}^{- 1})$ in $100 mL$ of water at $27^{\circ} C$ . The osmotic pressure of the solution is $(R = 0.08206 L atm K {mol}^{- 1})$

1

8.2 atm

2

2.46 atm

3

4.92 atm

4

1.64 atm

Explanation:

From formula,
$Π_{1} = C_{1} RT$
$Π_{2} = C_{2} RT$
Hence,
$\begin{aligned} Π = Π_{1} + Π_{2} \\ = & \frac{0.6 / 60}{0.1} RT + \frac{1.8 / 180}{0.1} RT \\ Π & = 2 (0.1) RT \\ = 0.2 \times 24.6 \\ = & 4.92 atm . \end{aligned}$

[JEE Main 2019

SOLUTIONS

277446 At $37^{\circ} C$ osmotic pressure of human blood is 7.65 atm. Tell how much glucose can be used in 1 lit of water for intravenous injection so that osmotic pressure of this glucose solution becomes equal to osmotic pressure of human blood.

1

22.2 gm

2

54.2 gm

3

15 gm

4

59.8 gm

Explanation:

Given that- $Π = 7.65$
$\begin{aligned} R = 0.08206 \\ T = 273 + 37 = 300 K \end{aligned}$
We know that-
$\begin{matrix} Π = CRT \\ 7.65 = C \times 0.08206 \times (273 + 37) \\ C = 0.3 M \end{matrix}$
Here, $1 L$ of solution will contain 0.3 moles of glucose
$= 0.3 \times 180 = 54 g$

Tripura JEE-2021

SOLUTIONS

277448 The osmotic pressure of a dilute solution of an ionic compound $XY$ in water is four times that of a solution of $0.01 M {BaCl}_{2}$ in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of $X Y$ (in mol $L^{- 1}$ ) in solution is

1

4 \times 10^{- 2}

2

16 \times 10^{- 4}

3

4 \times 10^{- 4}

4

6 \times 10^{- 2}

Explanation:

For xy
$i = 2 as complete dissociation takes place.$
$Π xy = 2 \times CRT$
For ${BaCl}_{2}$
$i = 3, Π {BaCl}_{2} = 3 \times 0.01 \times RT = 0.03 RT$
As $\prod xy = 4 \times \prod {BaCl}_{2}$
$\begin{aligned} \frac{\prod x y}{\prod {BaCl}_{2}} = \frac{2 \times CRT}{0.03 RT} \\ \frac{4 \prod {BaCl}_{2}}{\prod {BaCl}_{2}} = \frac{2 \times C}{0.03} \\ C = \frac{4 \times 0.03}{2} = 6 \times 10^{- 2} {molL}^{- 1} \end{aligned}$

[JEE Main 2019

SOLUTIONS

277449 Molal depression constant for a solvent is $4.0 K$ $kg mol^{- 1}$ . The depression in the freezing point of the solvent for $0.03 mol kg^{- 1}$ solution of $K_{2} {SO}_{4}$ is (Assume complete dissociation of the electrolyte)

1

0.18 K

2

0.36 K

3

0.12 K

4

0.24 K

Explanation:

Ionic Solution-
$K_{2} {SO}_{4} \to 2 K^{+} + {SO}^{2 -}$
Here $i = 3, m = 0.03 mol {kg}^{- 1}$
$Δ T_{f} = i \times k_{f} \times m$
$= 3 \times 4 \times 0.03$
$= 0.36 K$

[JEE Main 2019

SOLUTIONS

277450 A solution is prepared by dissolving $0.6 g$ of urea (molar mass $= 60 g {mol}^{- 1}$ ) and $1.8 g$ of glucose (molar mass $= 180 g {mol}^{- 1})$ in $100 mL$ of water at $27^{\circ} C$ . The osmotic pressure of the solution is $(R = 0.08206 L atm K {mol}^{- 1})$

1

8.2 atm

2

2.46 atm

3

4.92 atm

4

1.64 atm

Explanation:

From formula,
$Π_{1} = C_{1} RT$
$Π_{2} = C_{2} RT$
Hence,
$\begin{aligned} Π = Π_{1} + Π_{2} \\ = & \frac{0.6 / 60}{0.1} RT + \frac{1.8 / 180}{0.1} RT \\ Π & = 2 (0.1) RT \\ = 0.2 \times 24.6 \\ = & 4.92 atm . \end{aligned}$

[JEE Main 2019

SOLUTIONS

277446 At $37^{\circ} C$ osmotic pressure of human blood is 7.65 atm. Tell how much glucose can be used in 1 lit of water for intravenous injection so that osmotic pressure of this glucose solution becomes equal to osmotic pressure of human blood.

1

22.2 gm

2

54.2 gm

3

15 gm

4

59.8 gm

Explanation:

Given that- $Π = 7.65$
$\begin{aligned} R = 0.08206 \\ T = 273 + 37 = 300 K \end{aligned}$
We know that-
$\begin{matrix} Π = CRT \\ 7.65 = C \times 0.08206 \times (273 + 37) \\ C = 0.3 M \end{matrix}$
Here, $1 L$ of solution will contain 0.3 moles of glucose
$= 0.3 \times 180 = 54 g$

Tripura JEE-2021

SOLUTIONS

277448 The osmotic pressure of a dilute solution of an ionic compound $XY$ in water is four times that of a solution of $0.01 M {BaCl}_{2}$ in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of $X Y$ (in mol $L^{- 1}$ ) in solution is

1

4 \times 10^{- 2}

2

16 \times 10^{- 4}

3

4 \times 10^{- 4}

4

6 \times 10^{- 2}

Explanation:

For xy
$i = 2 as complete dissociation takes place.$
$Π xy = 2 \times CRT$
For ${BaCl}_{2}$
$i = 3, Π {BaCl}_{2} = 3 \times 0.01 \times RT = 0.03 RT$
As $\prod xy = 4 \times \prod {BaCl}_{2}$
$\begin{aligned} \frac{\prod x y}{\prod {BaCl}_{2}} = \frac{2 \times CRT}{0.03 RT} \\ \frac{4 \prod {BaCl}_{2}}{\prod {BaCl}_{2}} = \frac{2 \times C}{0.03} \\ C = \frac{4 \times 0.03}{2} = 6 \times 10^{- 2} {molL}^{- 1} \end{aligned}$

[JEE Main 2019

SOLUTIONS

277449 Molal depression constant for a solvent is $4.0 K$ $kg mol^{- 1}$ . The depression in the freezing point of the solvent for $0.03 mol kg^{- 1}$ solution of $K_{2} {SO}_{4}$ is (Assume complete dissociation of the electrolyte)

1

0.18 K

2

0.36 K

3

0.12 K

4

0.24 K

Explanation:

Ionic Solution-
$K_{2} {SO}_{4} \to 2 K^{+} + {SO}^{2 -}$
Here $i = 3, m = 0.03 mol {kg}^{- 1}$
$Δ T_{f} = i \times k_{f} \times m$
$= 3 \times 4 \times 0.03$
$= 0.36 K$

[JEE Main 2019

SOLUTIONS

277450 A solution is prepared by dissolving $0.6 g$ of urea (molar mass $= 60 g {mol}^{- 1}$ ) and $1.8 g$ of glucose (molar mass $= 180 g {mol}^{- 1})$ in $100 mL$ of water at $27^{\circ} C$ . The osmotic pressure of the solution is $(R = 0.08206 L atm K {mol}^{- 1})$

1

8.2 atm

2

2.46 atm

3

4.92 atm

4

1.64 atm

Explanation:

From formula,
$Π_{1} = C_{1} RT$
$Π_{2} = C_{2} RT$
Hence,
$\begin{aligned} Π = Π_{1} + Π_{2} \\ = & \frac{0.6 / 60}{0.1} RT + \frac{1.8 / 180}{0.1} RT \\ Π & = 2 (0.1) RT \\ = 0.2 \times 24.6 \\ = & 4.92 atm . \end{aligned}$

[JEE Main 2019