Question

Henry's law constant for $C{O}_2$ in water is $1.678$*${10}^8$pa at 298K the quantity of $C{O}_2$ in 500ml of soda water when packed under 2.5 atm pressure is :

• A. $14.9$ g
  
• B. $0.00084$ g
  
• C. $1.8$ g
  
• D. $8.4$ g
  

Answer 1

Answer: (C)

$1.8$ g

Pressure of ${CO}_2=25$ atm

$=2.5\times 1.013\times {10}^{13}{P}_a=2.533125\times {10}^5{P}_a$

$K_H=1.67\times {10}^8{P}_a$ (given)

According to Henny's law,

$P=\left(K_H\right)\left(X\right)$

$X=P/K_H=2.53\times {10}^5/1.67\times {10}^8=1.52\times {10}^{-3}$

But,

We have $500$ ml of soda water so that

Volume of water $=500$ ml

density of water $=1g/ml$

use formula mass $=$ volume $\times$ density

$500$ ml water $=500$g of water

Molar mass of water $\left(H_{2}O\right)=18g$ ${mol}^{-1}$

Number of moles $=\frac{mass}{molarmass}$

Number of mass of water $=\frac{500}{18}=27.78$ mole of water

mole function, $x=\frac{n_{{CO}_2}}{n_{{CO}_2}+n_{H_{2}O}}\approx \frac{n_{{CO}_2}}{n_{H_{2}O}}$

$1.52\times {10}^{-3}=\frac{n_{{CO}_2}}{27.78}$

$n_{{CO}_2}=0.042$ moles

molar mass of ${CO}_2=12+2\left(16\right)=44$ g/mol

mass $=$ (molar mass) (number of moles)

$=\left(0.042\right)\left(44\right)$

$=1.8$ grams

$\therefore$ Quantity of ${CO}_2$ is $1.8$ grams