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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

## 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. $$\frac{0.032}{20} \times 1000$$
$$= 1.60$$