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Question: How much oxygen is dissolved in 100 mL water at 298 K if partial pressure of oxygen is \(0.5\)atm an...

How much oxygen is dissolved in 100 mL water at 298 K if partial pressure of oxygen is 0.50.5atm and KH=1.4×10−3K _ { H } = 1.4 \times 10 ^ { - 3 }mol/ L/ atm

A

B

C

2.24 g

D

2.24 mg

Answer

2.24 mg

Explanation

Solution

According to Henry’s law where s is

concentration of O2\mathrm { O } _ { 2 }dissolved.

s=1.4×10−3×0.5=0.7×10−3 mol/L\mathrm { s } = 1.4 \times 10 ^ { - 3 } \times 0.5 = 0.7 \times 10 ^ { - 3 } \mathrm {~mol} / \mathrm { L }

s=nV\mathrm { s } = \frac { \mathrm { n } } { \mathrm { V } } or n=0.7×10−3×0.1=0.7×10−4 mol\mathrm { n } = 0.7 \times 10 ^ { - 3 } \times 0.1 = 0.7 \times 10 ^ { - 4 } \mathrm {~mol}

n=wM\mathrm { n } = \frac { \mathrm { w } } { \mathrm { M } } or w=n×M\mathrm { w } = \mathrm { n } \times \mathrm { M }

=0.7×10−4×32=22.4×10−4 g= 0.7 \times 10 ^ { - 4 } \times 32 = 22.4 \times 10 ^ { - 4 } \mathrm {~g} or 2.24 mg