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
jamb 2014chemistry

The correct answer is D. 0.50g

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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