If 10.5g of lead (II) trioxonitrate (V) is dissolved in 20cm³ of distilled water at 18^oC, the solubility of the solute in mol dm^-3 is
[Pb = 207, N = 14, O = 16]

tyjesma

11 years ago

Solubility=n • StndVol/volume...n=m/Mm..therefore sol.=m x stndvol/Mm x Volume....m=10.5, Mm={ Pb(NO3)2 }=331,Volume=20cm³, stnd volume=1dm³ ----»1000cm³.....Sol=10.5 x 1000/331 x 20 =1.59 ≈1.60

Samuel roseline

10 years ago

How did u get 331

Jay -jay. west

11 years ago

I need u guys 4 assistance

Anela

1 year ago

Sol.= m/mm × 1000/vol.
m= 10.5
mm of Pb(No3)2 = 207+2(

DAMMYSOUL

12 years ago

amnt of dissolved solute/molar mass of solute multiply by 1000/vol in cm3

Stock

10 years ago

Thanks