1.

When N_2 gas is passed through water at 293 K, how many moles of1N_2 woulddissolve in one litre water ? Assume that N_2 exert a partial pressure of 0.987 bar. K_H for N_2 at 293 K is 76.48 kilobar.

Answer»

Solution :By Henry.s law, MOLE fraction of `N_2 = P_(N_2)/K_H` =(0.987 bar)/(76480 bar) =`1.29xx10^(-5)`. One LITER water CONTAIN 55.5 mols of it. Let .n. be the number of mols of NITROGEN in solution , then,
`X_(N_2)= n/(n+55.5)= 1.29xx10^(-5)`
`therefore=1.29xx10^(-5)xx55.5=7.16xx10^(-4)`mol


Discussion

No Comment Found