Question

At 25C and 795 torr, carbon dioxide has a solubility of 0.0356 M in water. What is the solubility at 25 C and 2070 torr? ____ M

Answer 1

Answer:

0.09269 M is the solubility at 25°C and 2070 Torr.

Explanation:

Solubility of carbon dioxide in water at 25°C and 795 Torr = 0.0356 M

That is 0.0356 mol of carbon dioxide in 1 L of water.

Mass of 0.0356 moles of carbon dioxide =  0.0356mol × 44 g/mol =1.5664 g

Solubility of carbon dioxide in water at 25°C and 2070 Torr =?

According to Henry's law that is mass of gas dissolved in liquid at constant temperatures is directly proportional to the pressure of the gas at equilibrium with the liquid.

$m\propto Kp$

$\frac{m_1}{p_1}=\frac{m_2}{p_2}$

$m_1=1.5664 g,m_2=?$

$p_1=795 Torr,p_2=2070 Torr$

$m_2=\frac{1.5664\times 2070 Torr}{795 Torr}=4.0785 g$

Moles of carbon dioxide in 4.0785 g:$\frac{4.0785 g}{44 g/mol}=0.09269 mol$

0.09269 mol in 1 L of water.

So, the concentration of carbon dioxide will be 0.09269 M.

0.09269 M is the solubility at 25°C and 2070 Torr.

Answer 2

If you want to do real scientifically, the equation is 
p = kc. 
We evaluate k = p/c and plug in the first set of conditions. 
k = 795/0.0356 = 22,331 

Then p = kc and solve for c 
c = p/k = 2070/22,331 = 0.093