GENERAL,ORGANIC+BIOCHEM (LOOSELEAF)
GENERAL,ORGANIC+BIOCHEM (LOOSELEAF)
10th Edition
ISBN: 9781264035090
Author: Denniston
Publisher: MCG
Question
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Chapter 6, Problem 5MCP

(a)

Interpretation Introduction

Interpretation:

%(m/V) of 5gMgCl2 in 2L solution has to be calculated.

Concept introduction:

%(m/V) can be calculated using the following equation.

    %(m/V)=massofsolute(g)volumeofsolution(mL)×100%

(a)

Expert Solution
Check Mark

Answer to Problem 5MCP

%(m/V) of 5gMgCl2 in 2L(2000mL) solution is 0.25%.

Explanation of Solution

%(m/V) of 5gMgCl2 in 2L(2000mL) solution can be calculated as follows,

    %(m/V)=massofsolute(g)volumeofsolution(mL)×100%=5g2000mL×100%=0.25%.

Percent mass/volume is 0.25%.

(b)

Interpretation Introduction

Interpretation:

%(m/m) of 5gMgCl2 in 2L solution has to be calculated.

Concept introduction:

%(m/m) can be calculated using the following equation.

    %(m/m)=massofsolute(g)massofsolution(g)×100%

(b)

Expert Solution
Check Mark

Answer to Problem 5MCP

%(m/V) of 5gMgCl2 in 1.02g/L solution is 245%.

Explanation of Solution

Given that, the volume and density of solution are 2L and 1.02g/L respectively.

So, mass of solution can be calculated as follows,

    massofsolution=volume×density=2L×(1.02g/L)=2.04g

%(m/m) of 5gMgCl2 in 2.04g solution can be calculated as follows,

    %(m/m)=massofsolute(g)massofsolution(g)×100%=5g2.04g×100%=245%.

Percent mass/mass is 245%.

(c)

Interpretation Introduction

Interpretation:

Molarity of 5gMgCl2 in 2L solution has to be calculated.

Concept introduction:

Molarity can be calculated as follows,

    Molarity=moleofsoluteVolumeofsolution(L)

(c)

Expert Solution
Check Mark

Answer to Problem 5MCP

Molarity of 5gMgCl2 in 2L solution is 0.025M.

Explanation of Solution

Number of moles of magnesium chloride can be calculated as follows,

  NumberofmolesofMgCl2=givenmassmolarmass=5g95.211g/mole=0.05mole.

Therefore molarity of magnesium chloride solution is given below.

    Molarity=moleofsoluteVolumeofsolution(L)=0.05mole2L=0.025M.

Molarity is 0.025M.

(d)

Interpretation Introduction

Interpretation:

Molarity of magnesium ion 5gMgCl2 in 2L solution has to be calculated.

(d)

Expert Solution
Check Mark

Answer to Problem 5MCP

Molarity of magnesium ion 5gMgCl2 in 2L solution is 0.025M.

Explanation of Solution

Number of moles of magnesium chloride can be calculated as follows,

  NumberofmolesofMgCl2=givenmassmolarmass=5g95.211g/mole=0.05mole.

1mole Magnesium chloride contains 1mole magnesium ions.  So 0.05mole mole of magnesium chloride contains 0.05mole magnesium ions. 

Therefore molarity of magnesium ion in solution is given below.

    Molarity=moleofsoluteVolumeofsolution(L)=0.05mole2L=0.025M.

Molarity of magnesium ion is 0.025M.

(e)

Interpretation Introduction

Interpretation:

Molarity of chloride ion 5gMgCl2 in 2L solution has to be calculated.

(e)

Expert Solution
Check Mark

Answer to Problem 5MCP

Molarity of chloride ion 5gMgCl2 in 2L solution is 0.05M.

Explanation of Solution

Number of moles of magnesium chloride can be calculated as follows,

  NumberofmolesofMgCl2=givenmassmolarmass=5g95.211g/mole=0.05mole.

1mole Magnesium chloride contains 2mole chloride ions.  So 0.05mole mole of magnesium chloride contains 0.1mole chloride ions. 

Therefore molarity of chloride ion in solution is given below.

    Molarity=moleofsoluteVolumeofsolution(L)=0.1mole2L=0.05M.

Molarity of chloride ion is 0.05M.

(f)

Interpretation Introduction

Interpretation:

Osmotic pressure of 5gMgCl2 in 2L solution has to be calculated.

Concept introduction:

Osmotic pressure (π) can be calculated as follows,

  π=iMRT

Where,

    i is number of particles per mole of solute

    M is molar concentration of the solute

    R is universal gas constant

    T is temperature in Kelvin scale.

(f)

Expert Solution
Check Mark

Answer to Problem 5MCP

Osmotic pressure of 5gMgCl2 in 2L solution is 1.8atm.

Explanation of Solution

Molarity of magnesium chloride solution is 0.025M.

Given temperature is 25oC(298.15K).

The value of universal gas constant is 0.0821L.atm/K.mole.

There will be three ions per mole of the solution, because magnesium chloride dissociates to give one magnesium ion and two chloride ions.

Therefore osmotic pressure can be calculated as follows,

    π=iMRT=(3)×(0.025M)×(0.0821L.atm/K.mole.)×(298.15K)=1.8atm.

Osmotic pressure is 1.8atm.

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Chapter 6 Solutions

GENERAL,ORGANIC+BIOCHEM (LOOSELEAF)

Ch. 6.3 - Prob. 6.5PPCh. 6.3 - Prob. 6.6PPCh. 6.3 - Prob. 6.7PPCh. 6.3 - Prob. 6.7QCh. 6.3 - Prob. 6.8QCh. 6.3 - Prob. 6.8PPCh. 6.3 - Prob. 6.9PPCh. 6.4 - Prob. 6.9QCh. 6.4 - Comparing pure water and a 0.10 m glucose...Ch. 6.4 - Prob. 6.10PPCh. 6.4 - Prob. 6.11PPCh. 6.4 - Prob. 6.12PPCh. 6.4 - Prob. 6.13PPCh. 6.4 - Prob. 6.11QCh. 6.4 - Prob. 6.12QCh. 6.5 - Prob. 6.14PPCh. 6.5 - Prob. 6.15PPCh. 6.5 - Prob. 6.16PPCh. 6.5 - Prob. 6.13QCh. 6.5 - Prob. 6.14QCh. 6 - Prob. 6.15QPCh. 6 - Prob. 6.16QPCh. 6 - Which of the following solute(s) would form an...Ch. 6 - Prob. 6.18QPCh. 6 - Prob. 6.19QPCh. 6 - Prob. 6.20QPCh. 6 - Prob. 6.21QPCh. 6 - Prob. 6.22QPCh. 6 - Prob. 6.23QPCh. 6 - Prob. 6.24QPCh. 6 - Prob. 6.25QPCh. 6 - Prob. 6.26QPCh. 6 - Prob. 6.27QPCh. 6 - Prob. 6.28QPCh. 6 - The Henry’s law constant, k, for O2 in aqueous...Ch. 6 - The Henry’s law constant, k, for N2 in aqueous...Ch. 6 - Calculate the composition of each of the following...Ch. 6 - Calculate the composition of each of the following...Ch. 6 - Calculate the composition of each of the following...Ch. 6 - Calculate the composition of each of the following...Ch. 6 - Prob. 6.35QPCh. 6 - Calculate the composition of each of the following...Ch. 6 - Prob. 6.37QPCh. 6 - Prob. 6.38QPCh. 6 - Prob. 6.39QPCh. 6 - Prob. 6.40QPCh. 6 - Prob. 6.41QPCh. 6 - Prob. 6.42QPCh. 6 - Prob. 6.43QPCh. 6 - Prob. 6.44QPCh. 6 - Which solution is more concentrated: a 0.04% (m/m)...Ch. 6 - Which solution is more concentrated: a 20 ppt...Ch. 6 - Prob. 6.47QPCh. 6 - Prob. 6.48QPCh. 6 - Prob. 6.49QPCh. 6 - Prob. 6.50QPCh. 6 - Why is it often necessary to dilute solutions in...Ch. 6 - Write the dilution expression and define each...Ch. 6 - Prob. 6.53QPCh. 6 - Prob. 6.54QPCh. 6 - Prob. 6.55QPCh. 6 - Prob. 6.56QPCh. 6 - Calculate the volume of a 0.500 M sucrose solution...Ch. 6 - Calculate the volume of a 1.00 × 10−2 M KOH...Ch. 6 - It is desired to prepare 0.500 L of a 0.100 M...Ch. 6 - A 50.0-mL sample of a 0.250 M sucrose solution was...Ch. 6 - A 50.0-mL portion of a stock solution was diluted...Ch. 6 - Prob. 6.62QPCh. 6 - Prob. 6.63QPCh. 6 - Prob. 6.64QPCh. 6 - Prob. 6.65QPCh. 6 - Prob. 6.66QPCh. 6 - Prob. 6.67QPCh. 6 - Prob. 6.68QPCh. 6 - Prob. 6.69QPCh. 6 - Prob. 6.70QPCh. 6 - Prob. 6.71QPCh. 6 - Prob. 6.72QPCh. 6 - Prob. 6.73QPCh. 6 - Prob. 6.74QPCh. 6 - Prob. 6.75QPCh. 6 - Prob. 6.76QPCh. 6 - Prob. 6.77QPCh. 6 - Prob. 6.78QPCh. 6 - Answer Questions 6.79–6.82 based on the following...Ch. 6 - Answer Questions 6.79–6.82 based on the following...Ch. 6 - Answer Questions 6.79–6.82 based on the following...Ch. 6 - Answer Questions 6.79–6.82 based on the following...Ch. 6 - Prob. 6.83QPCh. 6 - Prob. 6.84QPCh. 6 - Prob. 6.85QPCh. 6 - Prob. 6.86QPCh. 6 - Prob. 6.87QPCh. 6 - Prob. 6.88QPCh. 6 - Prob. 6.89QPCh. 6 - Prob. 6.90QPCh. 6 - Prob. 6.91QPCh. 6 - Prob. 6.92QPCh. 6 - Prob. 6.93QPCh. 6 - Prob. 6.94QPCh. 6 - Prob. 6.95QPCh. 6 - Name the two most important cations in biological...Ch. 6 - Prob. 6.97QPCh. 6 - Explain why a dialysis solution must have an...Ch. 6 - Prob. 6.99QPCh. 6 - Prob. 6.100QPCh. 6 - Prob. 6.101QPCh. 6 - Prob. 6.102QPCh. 6 - Prob. 6.103QPCh. 6 - What type of solute dissolves readily in benzene...Ch. 6 - Prob. 6.105QPCh. 6 - Prob. 6.106QPCh. 6 - Prob. 6.107QPCh. 6 - Prob. 6.108QPCh. 6 - Prob. 6.109QPCh. 6 - Prob. 6.110QPCh. 6 - Prob. 6.111QPCh. 6 - Prob. 6.112QPCh. 6 - Prob. 6.113QPCh. 6 - Prob. 6.114QPCh. 6 - Prob. 1MCPCh. 6 - Prob. 2MCPCh. 6 - Prob. 3MCPCh. 6 - Prob. 4MCPCh. 6 - Prob. 5MCPCh. 6 - Prob. 6MCPCh. 6 - Prob. 7MCPCh. 6 - Prob. 8MCPCh. 6 - Prob. 9MCPCh. 6 - Prob. 10MCP
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