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$"8.855 g/L"$
This is just asking you to calculate the new molar (i.e. molar solubility) at a different pressure, i.e. use the . (We'll end up converting it later back to
$bb(PV = nRT)$ where:
$P$ is the pressure in$"atm"$ .$V$ is the volume in$"L"$ .$n$ is the mols of gas in$"mol"$ s.$R = "0.082057 L"cdot"atm/mol"cdot"K"$ is the universal gas constant in the appropriate units.$T$ is the temperature in$"K"$ .
Solving for molar density, we get:
$n/V = P/(RT)$
Since we have two solubilities to consider, we must have two states.
$n_1/V_1 = P_1/(RT)$
$n_2/V_2 = P_2/(RT)$
Therefore, we have:
$RT = (P_1V_1)/(n_1) = (P_2V_2)/(n_2)$
Solving for the new molar density, we get:
$n_2/V_2 = n_1/V_1 ((P_2)/(P_1))$
At this point, we can convert the mass-based solubility to mols. We assume that the
$n_1/V_1 = (0.161 cancel("g CO"_2) xx "1 mol CO"_2/(44.009 cancel("g CO"_2)))/("0.100 L water")$
$=$ $"0.0366 mol/L"$
Therefore, the new solubility is:
$"0.0366 mol/L" xx ("5.50 atm")/("1.00 atm")$
$=$ $"0.2012 mol/L"$
In the original units, we have:
$(0.2012 cancel("mols CO"_2))/"L solution" xx "44.009 g CO"_2/cancel("mol CO"_2)$
$=$ $color(blue)("8.855 g/L")$
Hence, the solubility increased by a factor of
(Note that we could have simply used the initial mass-based solubility as it is and avoided converting to molar solubility. But it is a good exercise to use the molar mass for conversions, so I avoided skipping that step.)
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