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

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Answer 1

Question

Chapter 11, Problem 11.92QA

Interpretation Introduction

To calculate:

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

Expert Solution & Answer

Answer to Problem 11.92QA

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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I have a question about this problem involving mechanisms and drawing curved arrows for acids and bases. I know we need to identify the nucleophile and electrophile, but are there different types of reactions? For instance, what about Grignard reagents and other types that I might not be familiar with? Can you help me with this? I want to identify the names of the mechanisms for problems 1-14, such as Gilman reagents and others. Are they all the same? Also, could you rewrite it so I can better understand? The handwriting is pretty cluttered. Additionally, I need to label the nucleophile and electrophile, but my main concern is whether those reactions differ, like the "Brønsted-Lowry acid-base mechanism, Lewis acid-base mechanism, acid-catalyzed mechanisms, acid-catalyzed reactions, base-catalyzed reactions, nucleophilic substitution mechanisms (SN1 and SN2), elimination reactions (E1 and E2), organometallic mechanisms, and so forth."

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Answer 2

Question

Chapter 11, Problem 11.92QA

Interpretation Introduction

To calculate:

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

Expert Solution & Answer

Answer to Problem 11.92QA

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 11.92QA

Interpretation Introduction

To calculate:

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

Expert Solution & Answer

Answer to Problem 11.92QA

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 11.92QA

Interpretation Introduction

To calculate:

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

Expert Solution & Answer

Answer to Problem 11.92QA

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 11.92QA

Interpretation Introduction

To calculate:

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

Expert Solution & Answer

Answer to Problem 11.92QA

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 11.92QA

Interpretation Introduction

To calculate:

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

Expert Solution & Answer

Answer to Problem 11.92QA

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 11.92QA

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 11.92QA

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. 11.5VP 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

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 11.92QA

Explanation of Solution

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Chapter 11 Solutions

To calculate:

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