The osmotic pressure (π) of (a) NaCl solution, (b) MgCl 2 solution, (c) mixture

Question

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Answer 1

Question

Chapter 11, Problem 11.82QA

Interpretation Introduction

To calculate:

The osmotic pressure (π) of (a) NaCl solution, (b) MgCl₂ solution, (c) mixture

Expert Solution & Answer

Answer to Problem 11.82QA

Solution:

Osmotic pressures are as follows:

aNaCl: 0.120 atm

bMgCl2:0.260 atm

cMixture:0.360 atm

Explanation of Solution

1. Concept:

NaCl→Na++Cl-

NaCl will dissociate into two ion as per the above equation, so the value for van’t Hoff’s factor (i) = 2

MgCl2→Mg2++2Cl-

MgCl₂ will dissociate into a total of three ions as per the above equation, so the value for van’t Hoff’s factor (i) = 3

2. Formula:

i. π=i ×M ×R ×T

Where, π is the osmotic pressure in atmosphere, M is the molar concentration (molarity) of all the particles in any solution, R is the universal gas constant (0.08206L×atmmol×K), T is the absolute temperature (in Kelvin).

A mole is the SI unit of amount chemical substance. When writing units, it is written as “mol”.

ii. Molarity is defined as:

molarityM=moles of solute L of solution

3. Given:

i. volume of NaCl=100.0 mL

ii. molarity of NaCl=2.50 mM

iii. volume of MgCl2=80.0 mL

iv. molarity of MgCl2=3.60 mM

v. temperature T=200C

4. Calculations:

temperature in K=(20+273.15)=293.15 K

a. Osmotic pressure (π) for NaCl:

Converting given mill molar solution into Molar solution as follows:

M inmolL=2.50 mM×1M1000 nM=0.0025 M=0.0025molL

Calculating osmotic pressure for NaCl:

π=2×0.0025molL×0.08206L.atmmol.K×293.15 K =0.120 atm

b. Osmotic pressure for MgCl₂:

Converting given mill molar solution into Molar solution as follows:

M inmolL=3.60 mM×1M1000 nM=0.0036 M=0.0036molL

π=3×0.0036molL×0.08206L.atmmol.K×293.15 K= 0.260 atm

c. Osmotic pressure for mixture of solution:

volume of NaCl in L=100.0 mL×1L1000 mL=0.1 L

molarityM=moles of solute L of solution

0.0025 M=moles NaCl0.1 L

moles of NaCl=0.00025 mol

volume of MgCl2 in L=80.0 mL×1L1000 mL=0.08 L

molarityM=moles of solute L of solution

0.0036 M=mol MgCl20.08 L

moles of NaCl=0.000288 mol

Total moles of mixture=0.00025 mol+0.000288 mol=0.000538 mol

Total volume of mixture=0.1 L+0.08L=0.18 L

i=2+3=5

molarity of mixture=0.000538 mol0.18 L=0.00299molL

π=5×0.00299molL×0.08206L.atmmol.K×293.15 K

π=0.360 atm

Conclusion:

Osmotic pressures are as follows:

aNaCl: 0.120 atm

bMgCl2:0.260 atm

cMixture:0.360 atm

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Answer 2

Question

Chapter 11, Problem 11.82QA

Interpretation Introduction

To calculate:

The osmotic pressure (π) of (a) NaCl solution, (b) MgCl₂ solution, (c) mixture

Expert Solution & Answer

Answer to Problem 11.82QA

Solution:

Osmotic pressures are as follows:

aNaCl: 0.120 atm

bMgCl2:0.260 atm

cMixture:0.360 atm

Explanation of Solution

1. Concept:

NaCl→Na++Cl-

NaCl will dissociate into two ion as per the above equation, so the value for van’t Hoff’s factor (i) = 2

MgCl2→Mg2++2Cl-

MgCl₂ will dissociate into a total of three ions as per the above equation, so the value for van’t Hoff’s factor (i) = 3

2. Formula:

i. π=i ×M ×R ×T

Where, π is the osmotic pressure in atmosphere, M is the molar concentration (molarity) of all the particles in any solution, R is the universal gas constant (0.08206L×atmmol×K), T is the absolute temperature (in Kelvin).

A mole is the SI unit of amount chemical substance. When writing units, it is written as “mol”.

ii. Molarity is defined as:

molarityM=moles of solute L of solution

3. Given:

i. volume of NaCl=100.0 mL

ii. molarity of NaCl=2.50 mM

iii. volume of MgCl2=80.0 mL

iv. molarity of MgCl2=3.60 mM

v. temperature T=200C

4. Calculations:

temperature in K=(20+273.15)=293.15 K

a. Osmotic pressure (π) for NaCl:

Converting given mill molar solution into Molar solution as follows:

M inmolL=2.50 mM×1M1000 nM=0.0025 M=0.0025molL

Calculating osmotic pressure for NaCl:

π=2×0.0025molL×0.08206L.atmmol.K×293.15 K =0.120 atm

b. Osmotic pressure for MgCl₂:

Converting given mill molar solution into Molar solution as follows:

M inmolL=3.60 mM×1M1000 nM=0.0036 M=0.0036molL

π=3×0.0036molL×0.08206L.atmmol.K×293.15 K= 0.260 atm

c. Osmotic pressure for mixture of solution:

volume of NaCl in L=100.0 mL×1L1000 mL=0.1 L

molarityM=moles of solute L of solution

0.0025 M=moles NaCl0.1 L

moles of NaCl=0.00025 mol

volume of MgCl2 in L=80.0 mL×1L1000 mL=0.08 L

molarityM=moles of solute L of solution

0.0036 M=mol MgCl20.08 L

moles of NaCl=0.000288 mol

Total moles of mixture=0.00025 mol+0.000288 mol=0.000538 mol

Total volume of mixture=0.1 L+0.08L=0.18 L

i=2+3=5

molarity of mixture=0.000538 mol0.18 L=0.00299molL

π=5×0.00299molL×0.08206L.atmmol.K×293.15 K

π=0.360 atm

Conclusion:

Osmotic pressures are as follows:

aNaCl: 0.120 atm

bMgCl2:0.260 atm

cMixture:0.360 atm

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Answer 3

Question

Chapter 11, Problem 11.82QA

Interpretation Introduction

To calculate:

The osmotic pressure (π) of (a) NaCl solution, (b) MgCl₂ solution, (c) mixture

Expert Solution & Answer

Answer to Problem 11.82QA

Solution:

Osmotic pressures are as follows:

aNaCl: 0.120 atm

bMgCl2:0.260 atm

cMixture:0.360 atm

Explanation of Solution

1. Concept:

NaCl→Na++Cl-

NaCl will dissociate into two ion as per the above equation, so the value for van’t Hoff’s factor (i) = 2

MgCl2→Mg2++2Cl-

MgCl₂ will dissociate into a total of three ions as per the above equation, so the value for van’t Hoff’s factor (i) = 3

2. Formula:

i. π=i ×M ×R ×T

Where, π is the osmotic pressure in atmosphere, M is the molar concentration (molarity) of all the particles in any solution, R is the universal gas constant (0.08206L×atmmol×K), T is the absolute temperature (in Kelvin).

A mole is the SI unit of amount chemical substance. When writing units, it is written as “mol”.

ii. Molarity is defined as:

molarityM=moles of solute L of solution

3. Given:

i. volume of NaCl=100.0 mL

ii. molarity of NaCl=2.50 mM

iii. volume of MgCl2=80.0 mL

iv. molarity of MgCl2=3.60 mM

v. temperature T=200C

4. Calculations:

temperature in K=(20+273.15)=293.15 K

a. Osmotic pressure (π) for NaCl:

Converting given mill molar solution into Molar solution as follows:

M inmolL=2.50 mM×1M1000 nM=0.0025 M=0.0025molL

Calculating osmotic pressure for NaCl:

π=2×0.0025molL×0.08206L.atmmol.K×293.15 K =0.120 atm

b. Osmotic pressure for MgCl₂:

Converting given mill molar solution into Molar solution as follows:

M inmolL=3.60 mM×1M1000 nM=0.0036 M=0.0036molL

π=3×0.0036molL×0.08206L.atmmol.K×293.15 K= 0.260 atm

c. Osmotic pressure for mixture of solution:

volume of NaCl in L=100.0 mL×1L1000 mL=0.1 L

molarityM=moles of solute L of solution

0.0025 M=moles NaCl0.1 L

moles of NaCl=0.00025 mol

volume of MgCl2 in L=80.0 mL×1L1000 mL=0.08 L

molarityM=moles of solute L of solution

0.0036 M=mol MgCl20.08 L

moles of NaCl=0.000288 mol

Total moles of mixture=0.00025 mol+0.000288 mol=0.000538 mol

Total volume of mixture=0.1 L+0.08L=0.18 L

i=2+3=5

molarity of mixture=0.000538 mol0.18 L=0.00299molL

π=5×0.00299molL×0.08206L.atmmol.K×293.15 K

π=0.360 atm

Conclusion:

Osmotic pressures are as follows:

aNaCl: 0.120 atm

bMgCl2:0.260 atm

cMixture:0.360 atm

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Answer 4

Question

Chapter 11, Problem 11.82QA

Interpretation Introduction

To calculate:

The osmotic pressure (π) of (a) NaCl solution, (b) MgCl₂ solution, (c) mixture

Expert Solution & Answer

Answer to Problem 11.82QA

Solution:

Osmotic pressures are as follows:

aNaCl: 0.120 atm

bMgCl2:0.260 atm

cMixture:0.360 atm

Explanation of Solution

1. Concept:

NaCl→Na++Cl-

NaCl will dissociate into two ion as per the above equation, so the value for van’t Hoff’s factor (i) = 2

MgCl2→Mg2++2Cl-

MgCl₂ will dissociate into a total of three ions as per the above equation, so the value for van’t Hoff’s factor (i) = 3

2. Formula:

i. π=i ×M ×R ×T

Where, π is the osmotic pressure in atmosphere, M is the molar concentration (molarity) of all the particles in any solution, R is the universal gas constant (0.08206L×atmmol×K), T is the absolute temperature (in Kelvin).

A mole is the SI unit of amount chemical substance. When writing units, it is written as “mol”.

ii. Molarity is defined as:

molarityM=moles of solute L of solution

3. Given:

i. volume of NaCl=100.0 mL

ii. molarity of NaCl=2.50 mM

iii. volume of MgCl2=80.0 mL

iv. molarity of MgCl2=3.60 mM

v. temperature T=200C

4. Calculations:

temperature in K=(20+273.15)=293.15 K

a. Osmotic pressure (π) for NaCl:

Converting given mill molar solution into Molar solution as follows:

M inmolL=2.50 mM×1M1000 nM=0.0025 M=0.0025molL

Calculating osmotic pressure for NaCl:

π=2×0.0025molL×0.08206L.atmmol.K×293.15 K =0.120 atm

b. Osmotic pressure for MgCl₂:

Converting given mill molar solution into Molar solution as follows:

M inmolL=3.60 mM×1M1000 nM=0.0036 M=0.0036molL

π=3×0.0036molL×0.08206L.atmmol.K×293.15 K= 0.260 atm

c. Osmotic pressure for mixture of solution:

volume of NaCl in L=100.0 mL×1L1000 mL=0.1 L

molarityM=moles of solute L of solution

0.0025 M=moles NaCl0.1 L

moles of NaCl=0.00025 mol

volume of MgCl2 in L=80.0 mL×1L1000 mL=0.08 L

molarityM=moles of solute L of solution

0.0036 M=mol MgCl20.08 L

moles of NaCl=0.000288 mol

Total moles of mixture=0.00025 mol+0.000288 mol=0.000538 mol

Total volume of mixture=0.1 L+0.08L=0.18 L

i=2+3=5

molarity of mixture=0.000538 mol0.18 L=0.00299molL

π=5×0.00299molL×0.08206L.atmmol.K×293.15 K

π=0.360 atm

Conclusion:

Osmotic pressures are as follows:

aNaCl: 0.120 atm

bMgCl2:0.260 atm

cMixture:0.360 atm

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Answer 5

Chapter 11, Problem 11.82QA

Interpretation Introduction

To calculate:

The osmotic pressure (π) of (a) NaCl solution, (b) MgCl₂ solution, (c) mixture

Expert Solution & Answer

Answer to Problem 11.82QA

Solution:

Osmotic pressures are as follows:

aNaCl: 0.120 atm

bMgCl2:0.260 atm

cMixture:0.360 atm

Explanation of Solution

1. Concept:

NaCl→Na++Cl-

NaCl will dissociate into two ion as per the above equation, so the value for van’t Hoff’s factor (i) = 2

MgCl2→Mg2++2Cl-

MgCl₂ will dissociate into a total of three ions as per the above equation, so the value for van’t Hoff’s factor (i) = 3

2. Formula:

i. π=i ×M ×R ×T

Where, π is the osmotic pressure in atmosphere, M is the molar concentration (molarity) of all the particles in any solution, R is the universal gas constant (0.08206L×atmmol×K), T is the absolute temperature (in Kelvin).

A mole is the SI unit of amount chemical substance. When writing units, it is written as “mol”.

ii. Molarity is defined as:

molarityM=moles of solute L of solution

3. Given:

i. volume of NaCl=100.0 mL

ii. molarity of NaCl=2.50 mM

iii. volume of MgCl2=80.0 mL

iv. molarity of MgCl2=3.60 mM

v. temperature T=200C

4. Calculations:

temperature in K=(20+273.15)=293.15 K

a. Osmotic pressure (π) for NaCl:

Converting given mill molar solution into Molar solution as follows:

M inmolL=2.50 mM×1M1000 nM=0.0025 M=0.0025molL

Calculating osmotic pressure for NaCl:

π=2×0.0025molL×0.08206L.atmmol.K×293.15 K =0.120 atm

b. Osmotic pressure for MgCl₂:

Converting given mill molar solution into Molar solution as follows:

M inmolL=3.60 mM×1M1000 nM=0.0036 M=0.0036molL

π=3×0.0036molL×0.08206L.atmmol.K×293.15 K= 0.260 atm

c. Osmotic pressure for mixture of solution:

volume of NaCl in L=100.0 mL×1L1000 mL=0.1 L

molarityM=moles of solute L of solution

0.0025 M=moles NaCl0.1 L

moles of NaCl=0.00025 mol

volume of MgCl2 in L=80.0 mL×1L1000 mL=0.08 L

molarityM=moles of solute L of solution

0.0036 M=mol MgCl20.08 L

moles of NaCl=0.000288 mol

Total moles of mixture=0.00025 mol+0.000288 mol=0.000538 mol

Total volume of mixture=0.1 L+0.08L=0.18 L

i=2+3=5

molarity of mixture=0.000538 mol0.18 L=0.00299molL

π=5×0.00299molL×0.08206L.atmmol.K×293.15 K

π=0.360 atm

Conclusion:

Osmotic pressures are as follows:

aNaCl: 0.120 atm

bMgCl2:0.260 atm

cMixture:0.360 atm

Answer 6

Chapter 11, Problem 11.82QA

Interpretation Introduction

To calculate:

The osmotic pressure (π) of (a) NaCl solution, (b) MgCl₂ solution, (c) mixture

Expert Solution & Answer

Answer to Problem 11.82QA

Solution:

Osmotic pressures are as follows:

aNaCl: 0.120 atm

bMgCl2:0.260 atm

cMixture:0.360 atm

Explanation of Solution

1. Concept:

NaCl→Na++Cl-

NaCl will dissociate into two ion as per the above equation, so the value for van’t Hoff’s factor (i) = 2

MgCl2→Mg2++2Cl-

MgCl₂ will dissociate into a total of three ions as per the above equation, so the value for van’t Hoff’s factor (i) = 3

2. Formula:

i. π=i ×M ×R ×T

Where, π is the osmotic pressure in atmosphere, M is the molar concentration (molarity) of all the particles in any solution, R is the universal gas constant (0.08206L×atmmol×K), T is the absolute temperature (in Kelvin).

A mole is the SI unit of amount chemical substance. When writing units, it is written as “mol”.

ii. Molarity is defined as:

molarityM=moles of solute L of solution

3. Given:

i. volume of NaCl=100.0 mL

ii. molarity of NaCl=2.50 mM

iii. volume of MgCl2=80.0 mL

iv. molarity of MgCl2=3.60 mM

v. temperature T=200C

4. Calculations:

temperature in K=(20+273.15)=293.15 K

a. Osmotic pressure (π) for NaCl:

Converting given mill molar solution into Molar solution as follows:

M inmolL=2.50 mM×1M1000 nM=0.0025 M=0.0025molL

Calculating osmotic pressure for NaCl:

π=2×0.0025molL×0.08206L.atmmol.K×293.15 K =0.120 atm

b. Osmotic pressure for MgCl₂:

Converting given mill molar solution into Molar solution as follows:

M inmolL=3.60 mM×1M1000 nM=0.0036 M=0.0036molL

π=3×0.0036molL×0.08206L.atmmol.K×293.15 K= 0.260 atm

c. Osmotic pressure for mixture of solution:

volume of NaCl in L=100.0 mL×1L1000 mL=0.1 L

molarityM=moles of solute L of solution

0.0025 M=moles NaCl0.1 L

moles of NaCl=0.00025 mol

volume of MgCl2 in L=80.0 mL×1L1000 mL=0.08 L

molarityM=moles of solute L of solution

0.0036 M=mol MgCl20.08 L

moles of NaCl=0.000288 mol

Total moles of mixture=0.00025 mol+0.000288 mol=0.000538 mol

Total volume of mixture=0.1 L+0.08L=0.18 L

i=2+3=5

molarity of mixture=0.000538 mol0.18 L=0.00299molL

π=5×0.00299molL×0.08206L.atmmol.K×293.15 K

π=0.360 atm

Conclusion:

Osmotic pressures are as follows:

aNaCl: 0.120 atm

bMgCl2:0.260 atm

cMixture:0.360 atm

Answer 7

Chapter 11, Problem 11.82QA

Interpretation Introduction

To calculate:

The osmotic pressure (π) of (a) NaCl solution, (b) MgCl₂ solution, (c) mixture

Answer 8

Expert Solution & Answer

Answer to Problem 11.82QA

Solution:

Osmotic pressures are as follows:

aNaCl: 0.120 atm

bMgCl2:0.260 atm

cMixture:0.360 atm

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

1. Concept:

NaCl→Na++Cl-

NaCl will dissociate into two ion as per the above equation, so the value for van’t Hoff’s factor (i) = 2

MgCl2→Mg2++2Cl-

MgCl₂ will dissociate into a total of three ions as per the above equation, so the value for van’t Hoff’s factor (i) = 3

2. Formula:

i. π=i ×M ×R ×T

Where, π is the osmotic pressure in atmosphere, M is the molar concentration (molarity) of all the particles in any solution, R is the universal gas constant (0.08206L×atmmol×K), T is the absolute temperature (in Kelvin).

A mole is the SI unit of amount chemical substance. When writing units, it is written as “mol”.

ii. Molarity is defined as:

molarityM=moles of solute L of solution

3. Given:

i. volume of NaCl=100.0 mL

ii. molarity of NaCl=2.50 mM

iii. volume of MgCl2=80.0 mL

iv. molarity of MgCl2=3.60 mM

v. temperature T=200C

4. Calculations:

temperature in K=(20+273.15)=293.15 K

a. Osmotic pressure (π) for NaCl:

Converting given mill molar solution into Molar solution as follows:

M inmolL=2.50 mM×1M1000 nM=0.0025 M=0.0025molL

Calculating osmotic pressure for NaCl:

π=2×0.0025molL×0.08206L.atmmol.K×293.15 K =0.120 atm

b. Osmotic pressure for MgCl₂:

Converting given mill molar solution into Molar solution as follows:

M inmolL=3.60 mM×1M1000 nM=0.0036 M=0.0036molL

π=3×0.0036molL×0.08206L.atmmol.K×293.15 K= 0.260 atm

c. Osmotic pressure for mixture of solution:

volume of NaCl in L=100.0 mL×1L1000 mL=0.1 L

molarityM=moles of solute L of solution

0.0025 M=moles NaCl0.1 L

moles of NaCl=0.00025 mol

volume of MgCl2 in L=80.0 mL×1L1000 mL=0.08 L

molarityM=moles of solute L of solution

0.0036 M=mol MgCl20.08 L

moles of NaCl=0.000288 mol

Total moles of mixture=0.00025 mol+0.000288 mol=0.000538 mol

Total volume of mixture=0.1 L+0.08L=0.18 L

i=2+3=5

molarity of mixture=0.000538 mol0.18 L=0.00299molL

π=5×0.00299molL×0.08206L.atmmol.K×293.15 K

π=0.360 atm

Conclusion:

Osmotic pressures are as follows:

aNaCl: 0.120 atm

bMgCl2:0.260 atm

cMixture:0.360 atm

Answer 12

Ch. 11 - Prob. 11.1VP Ch. 11 - Prob. 11.2VP Ch. 11 - Prob. 11.3VP Ch. 11 - Prob. 11.4VP Ch. 11 - Prob. 11.6VP Ch. 11 - Prob. 11.7VP Ch. 11 - Prob. 11.8VP Ch. 11 - Prob. 11.9VP Ch. 11 - Prob. 11.10VP Ch. 11 - Prob. 11.11QA

The osmotic pressure (π) of (a) NaCl solution, (b) MgCl 2 solution, (c) mixture

To calculate: The osmotic pressure (π) of (a) NaCl solution, (b) MgCl2 solution, (c) mixture

Answer to Problem 11.82QA

Explanation of Solution

Want to see more full solutions like this?

Chapter 11 Solutions

To calculate:

The osmotic pressure (π) of (a) NaCl solution, (b) MgCl₂ solution, (c) mixture