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]
- 0.05 g
- 2.00 g
- 1.00 g
-
0.50g
The correct answer is: D
Explanation
molar mass of pb (N0 3) 2 = 331
Mole = \(\frac{16.55}{331}\)= 0.05 mole
∴0.05mole pb (No3)2 dissolved in 1000g H2O x = \(\frac{0.05\times1000}{100}\)= 0.5 mole /dm 3
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