Question

Henry’s law constant for CO2 in water is 1.67 × 10 8 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.

Answer 1

Answer :

Given,

$K_{H} = 1.67 \times 10^{8}$ Pa

Pressure of $Co_{2}$, ρ$co_{2}$ = 2.5 atm = 2.5 × 101325 Pa

ρ$co_{2}$ = $K_{H}$ × χ$co_{2}$

⇒ χ$co_{2}$ = $\frac{ρco_{2}}{K_{H}}$ = $\frac{2.5 × 101325 Pa}{1.67 × 10^{8} Pa}$

χ$co_{2}$ = 1.517 × $10^{-3}$

⇒ $^{η}co_{2}$ = $\frac{^{η}co_{2}}{^{η}CO_{2} + ^{η}H_{2}O}$

⇒ $\frac{^{η}co_{2}}{^{η}H_{2}O}$= 1.517 × $10^{-3}$

[ ∵ $^{η}CO_{2}$ < < < $^{η}H_{2}O$ ]

∴ It can be neglected in the denominator.

⇒ $\frac{\dfrac{^{η}CO_{2}}{\ 500}}{18}$ = 1.517 × $10^{-3}$

⇒ $^{η}co_{2}$ = $\frac{500}{18} \times 1.517 \times 10^{-3}$

= $\frac{7.585}{18} \times 10^{-1} = \frac{75.85}{18} \times 10^{-2}$

= 4.214 × $10^{-2}$ moles

= 42.14 × $10^{-3}$ moles

= 42.14 millimoles

Answer 2

Answer: The amount of $CO_2$ in the given soda water is 1.848 g.

Explanation:

We are given:

$K_H=1.67\times 10^8Pa$

$p_{CO_2}=2.5atm=(2.5\times 101325)Pa$  (Conversion factor: 1 atm = 101325 Pa)

Using equation given by Henry's Law, we get:

$p_{CO_2}=K_H\times \chi_{CO_2}$

Putting values in above equation, we get:

$2.5\times 101325=1.67\times 10^8\times \chi_{CO_2}\\\\\chi_{CO_2}=1.51\times 10^{-3}$

Mole fraction of carbon dioxide in soda water is written as:

$\chi_{CO_2}=\frac{n_{CO_2}}{n_{CO_2}+n_{H_2O}}$

As, amount of water is more than the amount of carbon dioxide. Thus, moles of carbon dioxide will be neglected in the denominator.

Thus, $\chi_{CO_2}=\frac{n_{CO_2}}{n_{H_2O}}$     ......(1)

Density of water = 1 g/mL

So, the volume of water will be equal to the mass of water = 500 g

$1g/mL=\frac{\text{mass of water}}{500mL}\\\\\text{Mass of water}=500g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(2)

Given mass of water = 500 g

Molar mass of water = 18 g/mol

$\text{Moles of water}=\frac{500g}{18g/mol}=27.78mol$

Putting values in equation 1, we get:

$1.51\times 10^{-3}=\frac{n_{CO_2}}{27.78mol}\\\\n_{CO_2}=0.042mol$

Now, calculating the mass of carbon dioxide using equation 2, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.042 mol

Putting values in equation 2, we get:

$0.042mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=1.848g$

Hence, the amount of $CO_2$ in the given soda water is 1.848 g.