Question

Hey friends Please answer me fast

Answer 1

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PH2=KHXH2

XH2=PH2kH

=1bar7.18×103bar

=1.404×10−5

Now,

nH2O in 1L=1000g18

=55.5

Therefore ,

nH255.55=1.404×10−5

nH2=1.79×10−4moles

Hence the solubility of H2 in water is 7.79×10−4mol/L

Answer 2

10th class ka hai na

please wait I solve step by step