Chapter 4, Problem 4B.1E

(a)

Interpretation Introduction

Interpretation:

An open vessel contains water in a laboratory measuring $5.0m\times 4.3m\times 2.2m$ at ${25}^{o}C$. The vapour pressure of water is $2.3kPa.$ thus, the mass of the water which would found in the air if there is no ventilation has to be determined.

Concept introduction:

The number of moles of any substance can be determined using the equation

$\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mole (g/mol).

$Mass=Volume\times density$

Ideal gas equation,

$\text{PV}\text{=}\text{nRT}\boxed{\begin{array}{l}\text{Where,}\\ \text{P}\text{:}\text{Pressure}\\ \text{V:}\text{Volume}\\ \text{n:}\text{Number}\text{of}\text{moles}\\ \text{R:}\text{Gas}\text{constant}\\ \text{T}\text{:}\text{Temperature}\end{array}}$

Explanation of Solution

Given,

An open vessel contains water in a laboratory measuring $5.0m\times 4.3m\times 2.2m$ at ${25}^{o}C$. The vapour pressure of water is $2.3kPa.$ thus, the mass of the water which would found in the air if there is no ventilation can be determined as follows,

Volume of vessel can be determined as follows,

$\begin{array}{c}\text{5}\text{.0}\text{m}\text{×}\text{4}\text{.3}\text{m×}\text{2}\text{.2}\text{m}=\text{47}\text{.3}{\text{m}}^{\text{3}}\\ =\text{47}\text{.3}{\text{m}}^{\text{3}}\text{×}{\left(\frac{\text{1}\text{dm}}{{\text{10}}^{\text{-1}}\text{m}}\right)}^{\text{3}}\text{×}\left(\frac{\text{1L}}{\text{1}{\text{dm}}^{\text{3}}}\right)\\ =\text{47}\text{.3}\text{×}{\text{10}}^{\text{3}}\text{L}\end{array}$

Moles of water in this volume under the given consitions can be determined using Ideal gas equation as follows,

$\begin{array}{c}\text{n}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{=}\frac{\text{0}\text{.02269}\text{atm}\text{×}\text{47}\text{.3}\text{×}{\text{10}}^{\text{3}}\text{L}}{\left(\text{0}\text{.08206}\text{L}\text{.atm}{\text{.mol}}^{\text{-1}}{\text{.K}}^{\text{-1}}\right)\left(\text{25+273}\text{.15}\right)}\\ \text{=}\text{43}\text{.86}\text{mol}\end{array}$

The mass of water can be determined by multiplying this amount of water with its molar mass.

Thus,

$\begin{array}{c}Themassofwater=Molesofwater\times molarmassofwater\\ =43.86mol\times 18.0g/mol\\ =789.48g\end{array}$

(b)

Interpretation Introduction

Interpretation:

An open vessel contains benzene in a laboratory measuring $5.0m\times 4.3m\times 2.2m$ at ${25}^{o}C$. The vapour pressure of water is $\text{10 }kPa.$ thus, the mass of the benzene which would found in the air if there is no ventilation has to be determined.

Concept introduction:

The number of moles of any substance can be determined using the equation

$\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mole (g/mol).

$Mass=Volume\times density$

Ideal gas equation,

$\text{PV}\text{=}\text{nRT}\boxed{\begin{array}{l}\text{Where,}\\ \text{P}\text{:}\text{Pressure}\\ \text{V:}\text{Volume}\\ \text{n:}\text{Number}\text{of}\text{moles}\\ \text{R:}\text{Gas}\text{constant}\\ \text{T}\text{:}\text{Temperature}\end{array}}$

Explanation of Solution

Given,

An open vessel contains benzene in a laboratory measuring $5.0m\times 4.3m\times 2.2m$ at ${25}^{o}C$. The vapour pressure of water is $\text{10 }kPa.$ thus, the mass of the benzene which would found in the air if there is no ventilation can be determined as follows,

Volume of vessel can be determined as follows,

$\begin{array}{c}\text{5}\text{.0}\text{m}\text{×}\text{4}\text{.3}\text{m×}\text{2}\text{.2}\text{m}=\text{47}\text{.3}{\text{m}}^{\text{3}}\\ =\text{47}\text{.3}{\text{m}}^{\text{3}}\text{×}{\left(\frac{\text{1}\text{dm}}{{\text{10}}^{\text{-1}}\text{m}}\right)}^{\text{3}}\text{×}\left(\frac{\text{1L}}{\text{1}{\text{dm}}^{\text{3}}}\right)\\ =\text{47}\text{.3}\text{×}{\text{10}}^{\text{3}}\text{L}\end{array}$

Moles of benzene in this volume under the given consitions can be determined using Ideal gas equation as follows,

$\begin{array}{c}\text{n}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{=}\frac{\text{0}\text{.0986923}\text{atm}\text{×}\text{47}\text{.3}\text{×}{\text{10}}^{\text{3}}\text{L}}{\left(\text{0}\text{.08206}\text{L}\text{.atm}{\text{.mol}}^{\text{-1}}{\text{.K}}^{\text{-1}}\right)\left(\text{25+273}\text{.15}\right)}\\ \text{=}\text{190}\text{.8}\text{mol}\end{array}$

The mass of benzene can be determined by multiplying this amount of benzene with its molar mass.

Thus,

$\begin{array}{c}Themassofbenzene=Molesofbenzene\times molarmassofbenzene\\ =\text{190}\text{.8}\text{mol}\times 78.11g/mol\\ =14903.38g\end{array}$

(c)

Interpretation Introduction

Interpretation:

An open vessel contains mercury in a laboratory measuring $5.0m\times 4.3m\times 2.2m$ at ${25}^{o}C$. The vapour pressure of mercury is $\text{0}\text{.30 }kPa.$ thus, the mass of the mercury which would found in the air if there is no ventilation has to be determined.

Concept introduction:

The number of moles of any substance can be determined using the equation

$\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mole (g/mol).

$Mass=Volume\times density$

Ideal gas equation,

$\text{PV}\text{=}\text{nRT}\boxed{\begin{array}{l}\text{Where,}\\ \text{P}\text{:}\text{Pressure}\\ \text{V:}\text{Volume}\\ \text{n:}\text{Number}\text{of}\text{moles}\\ \text{R:}\text{Gas}\text{constant}\\ \text{T}\text{:}\text{Temperature}\end{array}}$

Explanation of Solution

Given,

An open vessel contains mercury in a laboratory measuring $5.0m\times 4.3m\times 2.2m$ at ${25}^{o}C$. The vapour pressure of mercury is $\text{0}\text{.30 }kPa.$ thus, the mass of the mercury which would found in the air if there is no ventilation can be determined as follows,

Volume of vessel can be determined as follows,

$\begin{array}{c}\text{5}\text{.0}\text{m}\text{×}\text{4}\text{.3}\text{m×}\text{2}\text{.2}\text{m}=\text{47}\text{.3}{\text{m}}^{\text{3}}\\ =\text{47}\text{.3}{\text{m}}^{\text{3}}\text{×}{\left(\frac{\text{1}\text{dm}}{{\text{10}}^{\text{-1}}\text{m}}\right)}^{\text{3}}\text{×}\left(\frac{\text{1L}}{\text{1}{\text{dm}}^{\text{3}}}\right)\\ =\text{47}\text{.3}\text{×}{\text{10}}^{\text{3}}\text{L}\end{array}$

Moles of water in this volume under the given consitions can be determined using Ideal gas equation as follows,

$\begin{array}{c}\text{n}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{=}\frac{\text{0}\text{.00296}\text{atm}\text{×}\text{47}\text{.3}\text{×}{\text{10}}^{\text{3}}\text{L}}{\left(\text{0}\text{.08206}\text{L}\text{.atm}{\text{.mol}}^{\text{-1}}{\text{.K}}^{\text{-1}}\right)\left(\text{25+273}\text{.15}\right)}\\ \text{=}\text{5}\text{.722}\text{mol}\end{array}$

The mass of mercury can be determined by multiplying this amount of mercury with its molar mass.

Thus,

$\begin{array}{c}Themassofmercury=Molesofmercury\times molarmassofmercury\\ =5.722mol\times 200.59g/mol\\ =1147.77g\end{array}$