Asked by Joseph
At 25oC, the equilibrium pressure of ammonia vapour above a 0.500 M aqueous ammonia solution is 0.84 kPa. Calculate the Henry's law constant for ammonia in M/bar. Do not include units. Include 4 significant figures.
Answers
Answered by
bobpursley
so, you have 1/2 mole, per liter, at .84kPa.
so change kPa to bar, .84kpa*.001bar/100Pa=840*1e-5=.0084bar
constant(M/p)=.5/.0084=59.52
so change kPa to bar, .84kpa*.001bar/100Pa=840*1e-5=.0084bar
constant(M/p)=.5/.0084=59.52
Answered by
Joseph
ok thankyou