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If 10.5g of lead (II) trioxonitrate (V) is dissolved in 20cm3 of distilled water...


Question

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

Options

A) 1.60

B) 5.25

C) 16.00

D) 525.00


The correct answer is A.

Explanation:

Mass of Pb(NO\(_3\))\(_2\) = 10.5g
Volume of distilled water = 20cm\(^3\)
By calculating the number of moles of lead trioxonitrate
Number of mole of Pb(NO\(_3\))\(_2\) = \(\frac{\text{Mass}}{\text{Molar Mass}}\)
But molar mass of Pb(NO\(_3\))\(_2\) = \(207 + 2(14 + (16 \times 3) = 331\)
Therefore, number of moles = \(\frac{10.5}{331} = 0.032\) mole
The solubility of lead trioxonitrate = \(\frac{\text{number of moles of lead trioxonitrate}}{\text{Volume of distilled water}}\) multiplied by 1000 cm\(^3\)
i.e. \(\frac{0.032}{20} \times 1000\)
\(= 1.60\)

Explanation provided by Aderogba Ibraheem


More Past Questions:


Dicussion (1)

  • Mass of Pb(NO\(_3\))\(_2\) = 10.5g
    Volume of distilled water = 20cm\(^3\)
    By calculating the number of moles of lead trioxonitrate
    Number of mole of Pb(NO\(_3\))\(_2\) = \(\frac{\text{Mass}}{\text{Molar Mass}}\)
    But molar mass of Pb(NO\(_3\))\(_2\) = \(207 + 2(14 + (16 \times 3) = 331\)
    Therefore, number of moles = \(\frac{10.5}{331} = 0.032\) mole
    The solubility of lead trioxonitrate = \(\frac{\text{number of moles of lead trioxonitrate}}{\text{Volume of distilled water}}\) multiplied by 1000 cm\(^3\)
    i.e. \(\frac{0.032}{20} \times 1000\)
    \(= 1.60\)