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myrzilka [38]
4 years ago
7

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

Chemistry
2 answers:
Burka [1]4 years ago
7 0

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.

Leya [2.2K]4 years ago
4 0
If you want to do real scientifically, the equation is 
<span>p = kc. </span>
<span>We evaluate k = p/c and plug in the first set of conditions. </span>
<span>k = 795/0.0356 = 22,331 </span>

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