calculate the solubility of the salute at 25 degree c
Answers
Answer:
Application of Henry’s Law
At 20 °C, the concentration of dissolved oxygen in water exposed to gaseous oxygen at a partial pressure of 101.3 kPa (760 torr) is 1.38 × 10-3 mol L-1. Use Henry’s law to determine the solubility of oxygen when its partial pressure is 20.7 kPa (155 torr), the approximate pressure of oxygen in earth’s atmosphere.
Solution
According to Henry’s law, for an ideal solution the solubility, Cg, of a gas (1.38 × 10-3 mol L-1, in this case) is directly proportional to the pressure, Pg, of the undissolved gas above the solution (101.3 kPa, or 760 torr, in this case). Because we know both Cg and Pg, we can rearrange this expression to solve for k.
C
g
=
k
P
g
k
=
C
g
P
g
=
1.38
×
10
−
3
mol
L
−
1
101.3
kPa
=
1.36
×
10
−
5
mol
L
−
1
kPa
−
1
(
1.82
×
10
−
6
mol
L
−
1
torr
−
1
)
Now we can use k to find the solubility at the lower pressure.
\displaystyle {C}_{\text{g}}=k{P}_{\text{g}}C
g
=kP
g
\displaystyle \begin{array}{l}\\ 1.36\times {10}^{-5}\text{mol}{\text{L}}^{-1}{\text{kPa}}^{-1}\times 20.7\text{kPa}\\ \left(\text{or}1.82\times {10}^{-6}\text{mol}{\text{L}}^{-1}{\text{torr}}^{-1}\times 155\text{torr}\right)\\ =2.82\times {10}^{-4}\text{mol}{\text{L}}^{-1}\end{array}
1.36×10
−5
molL
−1
kPa
−1
×20.7kPa
(or1.82×10
−6
molL
−1
torr
−1
×155torr)
=2.82×10
−4
molL
−1
Note that various units may be used to express the quantities involved in these sorts of computations. Any combination of units that yield to the constraints of dimensional analysis are acceptable.
Check Your Learning
Exposing a 100.0 mL sample of water at 0 °C to an atmosphere containing a gaseous solute at 20.26 kPa (152 torr) resulted in the dissolution of 1.45 × 10-3 g of the solute. Use Henry’s law to determine the solubility of this gaseous solute when its pressure is 101.3 kPa (760 torr).
Answer: 7.25 × 10-3 g
Explanation:
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