16.55g of lead (ll) trioxonitrate (V) was dissolved in 100g of distilled water at 20oC, calculate the solubility of the solute in moldm-3
[Pb = 207, N = 14, O = 16]
a
0.05 g
b
2.00 g
c
1.00 g
d
0.50g
Explanation
Correct Option
dVideo Explanation
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Discussions (19)

dandano
7 years ago
Molar mass of Pb(No3)2 =331. Mole=16.55/331=0.05mole. 0.05mole Pb(No3)2 dissolved in 100g H2O. x mole Pb(No3)2 dissolved in 1000g H2O. x=0.05*1000/100= 0.5mole/dm3.

Najibutajo1
4 months ago
The chosen Answer is correct = 0.5 mol/dm^3
But the Answer should in dm^3 not in gram

Anela
1 year ago
solubility= m/mm x 1000/vol
m= 16.55
mm= Pb(No3)2
= 207+2(14+3×16)
= 207+2(14+48)
= 207+2(62)
= 207+124= 331
vol.= 100
=16.55/331 x 1000/100
= 16550/33100
= 0.5(ANS)




