Q12.117CP - Chemistry: Molecular Nature Of Matter And Change

Question

A 0.75-L bottle is cleaned, dried, and closed in a room where the air is${\mathbf{22}}^o\mathbf{C}$   and 44% relative humidity (that is, the water vapor in the air is 0.44 of the equilibrium vapor pressure at data-custom-editor="chemistry" ${\mathbf{22}}^o\mathbf{C}$ ). The bottle is brought outside and stored at. data-custom-editor="chemistry" $\mathbf{0}\mathbf{.}{\mathbf{0}}^o\mathbf{C}$

 

(a) What mass of liquid water condenses inside the bottle?

 

(b) Would liquid water condense at?data-custom-editor="chemistry" ${\mathbf{10}}^o\mathbf{C}$ (See Table 5.3, p. 207.)

Step-by-Step Solution

Answer

As calculated, the mole of the water molecule at temperature $\text{0°C}$, $\text{10°C}$,  $\text{22°C}$ is 1.5mole, 3mole, and 6 molesrespectively, having a mass of 27g, 54g, and 108g.

 

(a) The difference between the mass at temperature  and   is 81g. hence, the mass of liquid water that condenses inside the bottle is 81grams.

(b) Yes, liquid mass condenses because of the difference between the mass at temperature  and . The amount of mass that condenses is 54grams

Subpart (a)

As known,

Pressure at temperature$\text{0°C}$ is 4.6 torr.

Pressure at temperature  $\text{10°C}$ is  .

Pressure at temperature  $\text{22°C}$is 19.8 torr.

Volume of the water $\text{= 0.75\hspace{0.17em}L = 750\hspace{0.17em}mL}$ .

Gas Constant,  .$\text{R = 8.314 J/k/mole}$

Step 1: The mass of the water molecule at : 0°C

The temperature $\text{0°C = 273K}$

Hence, the number of moles of water at$\text{0°C}$ :

 $\begin{array}{c}\text{PV = nRT}\\ \text{4.6 × 750mL = n × 8.314 × 273}\\ \text{n = }\frac{\text{4.6 × 750mL}}{\text{8.314 × 273}}\\ \text{n = 1.5 mole}\end{array}$

 

 

 

Thereofre, the mass of the water molecule at :$\text{0°C}$

$\begin{array}{c}\text{Number\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}Moles = }\frac{\text{Mass}}{\text{Molar\hspace{0.17em}Mass}}\\ \text{1.5 = }\frac{\text{Mass}}{\text{18}}\end{array}$

So,

$\begin{array}{c}\text{Mass = 18 × 1.5}\\ \text{Mass = 27 g}\end{array}$ 

 

Step 2: The mass of the water molecule at: 22°C

Temperature  $\text{22°C = 295K}$

Hence, the number of moles of water at :$\text{22°C}$

 

 

 $\begin{array}{c}\text{PV = nRT}\\ \text{19.8 × 750mL = n × 8.314 × 295}\\ \text{n = }\frac{\text{19.8 × 750mL}}{\text{8.314 × 295}}\\ \text{n = 6 mole}\end{array}$

 

Thereofre, the mass of the water molecule at :$\text{22°C}$

 $\begin{array}{c}\text{Number\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}Moles = }\frac{\text{Mass}}{\text{Molar\hspace{0.17em}Mass}}\\ \text{6 = }\frac{\text{Mass}}{\text{18}}\end{array}$

 

So,

$\begin{array}{c}\text{Mass = 18 × 6}\\ \text{Mass = 108 g}\end{array}$ 

Step 3: Themass of liquid water condenses inside the bottle.

The difference in the mass at different temperature$\text{0°C}$ and$\text{22°C}$ :

 

 $\begin{array}{l}\text{ΔM = 108g - 27g}\\ \text{ΔM = 81g}\end{array}$

 

 

Subpart (b) Step 1: Mass of the water molecule at :

Temperature  $\text{10°C = 283K}$

So, the number of moles of water at :$\text{10°C}$

 

 $\begin{array}{c}\text{PV = nRT}\\ \text{9.2 × 750mL = n × 8.314 × 283}\\ \text{n = }\frac{\text{9.2 × 750mL}}{\text{8.314 × 283}}\\ \text{n = 3mole}\end{array}$

Hence, the mass of the water molecule at:$\text{10°C}$

$\begin{array}{c}\text{Number\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}Moles = }\frac{\text{Mass}}{\text{MolarMass}}\\ \text{3 = }\frac{\text{Mass}}{\text{18}}\end{array}$ 

Hence,

$\begin{array}{c}\text{Mass = 18 × 3}\\ \text{Mass = 54g}\end{array}$ 

Step 2:The mass of liquid water that condense at . 10°C

The difference in the mass at different temperature $\text{10°C}$and:$\text{22°C}$

 $\begin{array}{l}\text{ΔM = 108g - 54g}\\ \text{ΔM = 54g}\end{array}$