The molar mass of unknown gas with mass 2 .85 g , volume 750 mL at 760 mm Hg and 100 ° C is to be stated. Concept introduction: The combination of all three ideal gas laws: Boyle's Law, Charles' Law, and Gay-Lussac's Law gives rise to a combined gas law . According to this law, pressure, volume, temperature and number of moles of a gas are related to each other.

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Answer 1

Question

Chapter 10, Problem 84E

Interpretation Introduction

Interpretation:

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is to be stated.

Concept introduction:

The combination of all three ideal gas laws: Boyle's Law, Charles' Law, and Gay-Lussac's Law gives rise to a combined gas law. According to this law, pressure, volume, temperature and number of moles of a gas are related to each other.

Expert Solution & Answer

Answer to Problem 84E

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is $116.4 g/mol$ .

Explanation of Solution

The equation for the combined gas law is given below.

$PV=nRT$ …(1)

Where,

• $P$ is the pressure of unknown gas.

• $V$ is the volume of unknown gas.

• $n$ is the number of moles of unknown gas.

• $T$ is the temperature of unknown gas.

• $R$ is the gas constant.

• $V$ is the volume of unknown gas.

The given value of pressure is $760 mm Hg$ .

The given value of volume is $750 mL$ .

The given value of temperature is $100 °C$ .

The given value of mass is $2.85 g$ .

The given value of $R$ is $0.0821 L⋅atm/mol⋅K$ .

The formula of number of moles is shown below.

$n(mol)=Mass of unknown gas(g)Molar mass of unknown gas$ …(2)

Substitute the value of (2) in equation (1) as shown below.

$PV=nRTPV=(Mass of unknown gas(g)Molar mass of unknown gas)RT$ …(3)

Conversion of temperature into Kelvin is shown below.

$T=(T(°C)+273.15)K=(100+273.15) K=373.15 K$

Conversion of volume into liters is shown below.

$1000 mL=1 L1 mL=1 L1000 mL$

Value of $750 mL$ in $L$ is stated below.

$750 mL=1 L1000 mL×750 mL=0.75 L$

Conversion of $760 mm Hg$ pressure into $atm$ is shown below.

$760 mm Hg=1 atm$

The value of $P,V,R,T$ and mass of unknown gas is $1 atm,0.75 L,0.0821 L⋅atm/mol⋅K,373.15 K$ and $2.85 g$ respectively.

Substitute the values of $P,V,R,T$ and mass of unknown gas in the equation (3) as shown below.

$PV=(Mass of unknown gas(g)Molar mass of unknown gas)RT1 atm×0.75 L=(2.85 gMolar mass of unknown gas)×0.0821 L⋅atm/mol⋅K×373.15 KMolar mass of unknown gas=(2.85 g1 atm×0.75 L)×0.0821 L⋅atm/mol⋅K×373.15 KMolar mass of unknown gas=116.4 g/mol$

Conclusion

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is calculated as $116.4 g/mol$ .

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Answer 2

Question

Chapter 10, Problem 84E

Interpretation Introduction

Interpretation:

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is to be stated.

Concept introduction:

The combination of all three ideal gas laws: Boyle's Law, Charles' Law, and Gay-Lussac's Law gives rise to a combined gas law. According to this law, pressure, volume, temperature and number of moles of a gas are related to each other.

Expert Solution & Answer

Answer to Problem 84E

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is $116.4 g/mol$ .

Explanation of Solution

The equation for the combined gas law is given below.

$PV=nRT$ …(1)

Where,

• $P$ is the pressure of unknown gas.

• $V$ is the volume of unknown gas.

• $n$ is the number of moles of unknown gas.

• $T$ is the temperature of unknown gas.

• $R$ is the gas constant.

• $V$ is the volume of unknown gas.

The given value of pressure is $760 mm Hg$ .

The given value of volume is $750 mL$ .

The given value of temperature is $100 °C$ .

The given value of mass is $2.85 g$ .

The given value of $R$ is $0.0821 L⋅atm/mol⋅K$ .

The formula of number of moles is shown below.

$n(mol)=Mass of unknown gas(g)Molar mass of unknown gas$ …(2)

Substitute the value of (2) in equation (1) as shown below.

$PV=nRTPV=(Mass of unknown gas(g)Molar mass of unknown gas)RT$ …(3)

Conversion of temperature into Kelvin is shown below.

$T=(T(°C)+273.15)K=(100+273.15) K=373.15 K$

Conversion of volume into liters is shown below.

$1000 mL=1 L1 mL=1 L1000 mL$

Value of $750 mL$ in $L$ is stated below.

$750 mL=1 L1000 mL×750 mL=0.75 L$

Conversion of $760 mm Hg$ pressure into $atm$ is shown below.

$760 mm Hg=1 atm$

The value of $P,V,R,T$ and mass of unknown gas is $1 atm,0.75 L,0.0821 L⋅atm/mol⋅K,373.15 K$ and $2.85 g$ respectively.

Substitute the values of $P,V,R,T$ and mass of unknown gas in the equation (3) as shown below.

$PV=(Mass of unknown gas(g)Molar mass of unknown gas)RT1 atm×0.75 L=(2.85 gMolar mass of unknown gas)×0.0821 L⋅atm/mol⋅K×373.15 KMolar mass of unknown gas=(2.85 g1 atm×0.75 L)×0.0821 L⋅atm/mol⋅K×373.15 KMolar mass of unknown gas=116.4 g/mol$

Conclusion

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is calculated as $116.4 g/mol$ .

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Answer 3

Question

Chapter 10, Problem 84E

Interpretation Introduction

Interpretation:

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is to be stated.

Concept introduction:

The combination of all three ideal gas laws: Boyle's Law, Charles' Law, and Gay-Lussac's Law gives rise to a combined gas law. According to this law, pressure, volume, temperature and number of moles of a gas are related to each other.

Expert Solution & Answer

Answer to Problem 84E

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is $116.4 g/mol$ .

Explanation of Solution

The equation for the combined gas law is given below.

$PV=nRT$ …(1)

Where,

• $P$ is the pressure of unknown gas.

• $V$ is the volume of unknown gas.

• $n$ is the number of moles of unknown gas.

• $T$ is the temperature of unknown gas.

• $R$ is the gas constant.

• $V$ is the volume of unknown gas.

The given value of pressure is $760 mm Hg$ .

The given value of volume is $750 mL$ .

The given value of temperature is $100 °C$ .

The given value of mass is $2.85 g$ .

The given value of $R$ is $0.0821 L⋅atm/mol⋅K$ .

The formula of number of moles is shown below.

$n(mol)=Mass of unknown gas(g)Molar mass of unknown gas$ …(2)

Substitute the value of (2) in equation (1) as shown below.

$PV=nRTPV=(Mass of unknown gas(g)Molar mass of unknown gas)RT$ …(3)

Conversion of temperature into Kelvin is shown below.

$T=(T(°C)+273.15)K=(100+273.15) K=373.15 K$

Conversion of volume into liters is shown below.

$1000 mL=1 L1 mL=1 L1000 mL$

Value of $750 mL$ in $L$ is stated below.

$750 mL=1 L1000 mL×750 mL=0.75 L$

Conversion of $760 mm Hg$ pressure into $atm$ is shown below.

$760 mm Hg=1 atm$

The value of $P,V,R,T$ and mass of unknown gas is $1 atm,0.75 L,0.0821 L⋅atm/mol⋅K,373.15 K$ and $2.85 g$ respectively.

Substitute the values of $P,V,R,T$ and mass of unknown gas in the equation (3) as shown below.

$PV=(Mass of unknown gas(g)Molar mass of unknown gas)RT1 atm×0.75 L=(2.85 gMolar mass of unknown gas)×0.0821 L⋅atm/mol⋅K×373.15 KMolar mass of unknown gas=(2.85 g1 atm×0.75 L)×0.0821 L⋅atm/mol⋅K×373.15 KMolar mass of unknown gas=116.4 g/mol$

Conclusion

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is calculated as $116.4 g/mol$ .

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Answer 4

Question

Chapter 10, Problem 84E

Interpretation Introduction

Interpretation:

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is to be stated.

Concept introduction:

The combination of all three ideal gas laws: Boyle's Law, Charles' Law, and Gay-Lussac's Law gives rise to a combined gas law. According to this law, pressure, volume, temperature and number of moles of a gas are related to each other.

Expert Solution & Answer

Answer to Problem 84E

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is $116.4 g/mol$ .

Explanation of Solution

The equation for the combined gas law is given below.

$PV=nRT$ …(1)

Where,

• $P$ is the pressure of unknown gas.

• $V$ is the volume of unknown gas.

• $n$ is the number of moles of unknown gas.

• $T$ is the temperature of unknown gas.

• $R$ is the gas constant.

• $V$ is the volume of unknown gas.

The given value of pressure is $760 mm Hg$ .

The given value of volume is $750 mL$ .

The given value of temperature is $100 °C$ .

The given value of mass is $2.85 g$ .

The given value of $R$ is $0.0821 L⋅atm/mol⋅K$ .

The formula of number of moles is shown below.

$n(mol)=Mass of unknown gas(g)Molar mass of unknown gas$ …(2)

Substitute the value of (2) in equation (1) as shown below.

$PV=nRTPV=(Mass of unknown gas(g)Molar mass of unknown gas)RT$ …(3)

Conversion of temperature into Kelvin is shown below.

$T=(T(°C)+273.15)K=(100+273.15) K=373.15 K$

Conversion of volume into liters is shown below.

$1000 mL=1 L1 mL=1 L1000 mL$

Value of $750 mL$ in $L$ is stated below.

$750 mL=1 L1000 mL×750 mL=0.75 L$

Conversion of $760 mm Hg$ pressure into $atm$ is shown below.

$760 mm Hg=1 atm$

The value of $P,V,R,T$ and mass of unknown gas is $1 atm,0.75 L,0.0821 L⋅atm/mol⋅K,373.15 K$ and $2.85 g$ respectively.

Substitute the values of $P,V,R,T$ and mass of unknown gas in the equation (3) as shown below.

$PV=(Mass of unknown gas(g)Molar mass of unknown gas)RT1 atm×0.75 L=(2.85 gMolar mass of unknown gas)×0.0821 L⋅atm/mol⋅K×373.15 KMolar mass of unknown gas=(2.85 g1 atm×0.75 L)×0.0821 L⋅atm/mol⋅K×373.15 KMolar mass of unknown gas=116.4 g/mol$

Conclusion

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is calculated as $116.4 g/mol$ .

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Answer 5

Chapter 10, Problem 84E

Interpretation Introduction

Interpretation:

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is to be stated.

Concept introduction:

The combination of all three ideal gas laws: Boyle's Law, Charles' Law, and Gay-Lussac's Law gives rise to a combined gas law. According to this law, pressure, volume, temperature and number of moles of a gas are related to each other.

Expert Solution & Answer

Answer to Problem 84E

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is $116.4 g/mol$ .

Explanation of Solution

The equation for the combined gas law is given below.

$PV=nRT$ …(1)

Where,

• $P$ is the pressure of unknown gas.

• $V$ is the volume of unknown gas.

• $n$ is the number of moles of unknown gas.

• $T$ is the temperature of unknown gas.

• $R$ is the gas constant.

• $V$ is the volume of unknown gas.

The given value of pressure is $760 mm Hg$ .

The given value of volume is $750 mL$ .

The given value of temperature is $100 °C$ .

The given value of mass is $2.85 g$ .

The given value of $R$ is $0.0821 L⋅atm/mol⋅K$ .

The formula of number of moles is shown below.

$n(mol)=Mass of unknown gas(g)Molar mass of unknown gas$ …(2)

Substitute the value of (2) in equation (1) as shown below.

$PV=nRTPV=(Mass of unknown gas(g)Molar mass of unknown gas)RT$ …(3)

Conversion of temperature into Kelvin is shown below.

$T=(T(°C)+273.15)K=(100+273.15) K=373.15 K$

Conversion of volume into liters is shown below.

$1000 mL=1 L1 mL=1 L1000 mL$

Value of $750 mL$ in $L$ is stated below.

$750 mL=1 L1000 mL×750 mL=0.75 L$

Conversion of $760 mm Hg$ pressure into $atm$ is shown below.

$760 mm Hg=1 atm$

The value of $P,V,R,T$ and mass of unknown gas is $1 atm,0.75 L,0.0821 L⋅atm/mol⋅K,373.15 K$ and $2.85 g$ respectively.

Substitute the values of $P,V,R,T$ and mass of unknown gas in the equation (3) as shown below.

$PV=(Mass of unknown gas(g)Molar mass of unknown gas)RT1 atm×0.75 L=(2.85 gMolar mass of unknown gas)×0.0821 L⋅atm/mol⋅K×373.15 KMolar mass of unknown gas=(2.85 g1 atm×0.75 L)×0.0821 L⋅atm/mol⋅K×373.15 KMolar mass of unknown gas=116.4 g/mol$

Conclusion

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is calculated as $116.4 g/mol$ .

Answer 6

Chapter 10, Problem 84E

Interpretation Introduction

Interpretation:

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is to be stated.

Concept introduction:

The combination of all three ideal gas laws: Boyle's Law, Charles' Law, and Gay-Lussac's Law gives rise to a combined gas law. According to this law, pressure, volume, temperature and number of moles of a gas are related to each other.

Expert Solution & Answer

Answer to Problem 84E

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is $116.4 g/mol$ .

Explanation of Solution

The equation for the combined gas law is given below.

$PV=nRT$ …(1)

Where,

• $P$ is the pressure of unknown gas.

• $V$ is the volume of unknown gas.

• $n$ is the number of moles of unknown gas.

• $T$ is the temperature of unknown gas.

• $R$ is the gas constant.

• $V$ is the volume of unknown gas.

The given value of pressure is $760 mm Hg$ .

The given value of volume is $750 mL$ .

The given value of temperature is $100 °C$ .

The given value of mass is $2.85 g$ .

The given value of $R$ is $0.0821 L⋅atm/mol⋅K$ .

The formula of number of moles is shown below.

$n(mol)=Mass of unknown gas(g)Molar mass of unknown gas$ …(2)

Substitute the value of (2) in equation (1) as shown below.

$PV=nRTPV=(Mass of unknown gas(g)Molar mass of unknown gas)RT$ …(3)

Conversion of temperature into Kelvin is shown below.

$T=(T(°C)+273.15)K=(100+273.15) K=373.15 K$

Conversion of volume into liters is shown below.

$1000 mL=1 L1 mL=1 L1000 mL$

Value of $750 mL$ in $L$ is stated below.

$750 mL=1 L1000 mL×750 mL=0.75 L$

Conversion of $760 mm Hg$ pressure into $atm$ is shown below.

$760 mm Hg=1 atm$

The value of $P,V,R,T$ and mass of unknown gas is $1 atm,0.75 L,0.0821 L⋅atm/mol⋅K,373.15 K$ and $2.85 g$ respectively.

Substitute the values of $P,V,R,T$ and mass of unknown gas in the equation (3) as shown below.

$PV=(Mass of unknown gas(g)Molar mass of unknown gas)RT1 atm×0.75 L=(2.85 gMolar mass of unknown gas)×0.0821 L⋅atm/mol⋅K×373.15 KMolar mass of unknown gas=(2.85 g1 atm×0.75 L)×0.0821 L⋅atm/mol⋅K×373.15 KMolar mass of unknown gas=116.4 g/mol$

Conclusion

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is calculated as $116.4 g/mol$ .

Answer 7

Chapter 10, Problem 84E

Interpretation Introduction

Interpretation:

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is to be stated.

Concept introduction:

The combination of all three ideal gas laws: Boyle's Law, Charles' Law, and Gay-Lussac's Law gives rise to a combined gas law. According to this law, pressure, volume, temperature and number of moles of a gas are related to each other.

Answer 8

Expert Solution & Answer

Answer to Problem 84E

The molar mass of unknown gas with mass $2.85 g$ , volume $750 mL$ at $760 mm Hg$ and $100 °C$ is $116.4 g/mol$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The equation for the combined gas law is given below.

$PV=nRT$ …(1)

Where,

• $P$ is the pressure of unknown gas.

• $V$ is the volume of unknown gas.

• $n$ is the number of moles of unknown gas.

• $T$ is the temperature of unknown gas.

• $R$ is the gas constant.

• $V$ is the volume of unknown gas.

The given value of pressure is $760 mm Hg$ .

The given value of volume is $750 mL$ .

The given value of temperature is $100 °C$ .

The given value of mass is $2.85 g$ .

The given value of $R$ is $0.0821 L⋅atm/mol⋅K$ .

The formula of number of moles is shown below.

$n(mol)=Mass of unknown gas(g)Molar mass of unknown gas$ …(2)

Substitute the value of (2) in equation (1) as shown below.

$PV=nRTPV=(Mass of unknown gas(g)Molar mass of unknown gas)RT$ …(3)

Conversion of temperature into Kelvin is shown below.

$T=(T(°C)+273.15)K=(100+273.15) K=373.15 K$

Conversion of volume into liters is shown below.

$1000 mL=1 L1 mL=1 L1000 mL$

Value of $750 mL$ in $L$ is stated below.

$750 mL=1 L1000 mL×750 mL=0.75 L$

Conversion of $760 mm Hg$ pressure into $atm$ is shown below.

$760 mm Hg=1 atm$

The value of $P,V,R,T$ and mass of unknown gas is $1 atm,0.75 L,0.0821 L⋅atm/mol⋅K,373.15 K$ and $2.85 g$ respectively.

Substitute the values of $P,V,R,T$ and mass of unknown gas in the equation (3) as shown below.

$PV=(Mass of unknown gas(g)Molar mass of unknown gas)RT1 atm×0.75 L=(2.85 gMolar mass of unknown gas)×0.0821 L⋅atm/mol⋅K×373.15 KMolar mass of unknown gas=(2.85 g1 atm×0.75 L)×0.0821 L⋅atm/mol⋅K×373.15 KMolar mass of unknown gas=116.4 g/mol$