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If 60 cm3 of a saturated solution of KNO3 contains 15.6 g of the salt at 35 °C, calculate the solubility of KNO3 in mol dm-3 at that temperature. [K = 39.0, O = 16.0, N = 14.0]

Chemistry

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  1. Calculate the molar mass of KNO3\text{KNO}_3.

    M(KNO3)=M(K)+M(N)+3×M(O)M(\text{KNO}_3) = M(\text{K}) + M(\text{N}) + 3 \times M(\text{O}) M(KNO3)=39.0+14.0+(3×16.0)=39.0+14.0+48.0=101.0 g/molM(\text{KNO}_3) = 39.0 + 14.0 + (3 \times 16.0) = 39.0 + 14.0 + 48.0 = 101.0 \text{ g/mol}

  2. Calculate the number of moles of KNO3\text{KNO}_3.

    Moles of KNO3=Mass of KNO3Molar mass of KNO3\text{Moles of KNO}_3 = \frac{\text{Mass of KNO}_3}{\text{Molar mass of KNO}_3} Moles of KNO3=15.6 g101.0 g/mol≈0.15446 mol\text{Moles of KNO}_3 = \frac{15.6 \text{ g}}{101.0 \text{ g/mol}} \approx 0.15446 \text{ mol}

  3. Convert the volume of the solution from cm3\text{cm}^3 to dm3\text{dm}^3.

    1 dm3=1000 cm31 \text{ dm}^3 = 1000 \text{ cm}^3 Volume=60 cm3=601000 dm3=0.060 dm3\text{Volume} = 60 \text{ cm}^3 = \frac{60}{1000} \text{ dm}^3 = 0.060 \text{ dm}^3

  4. Calculate the solubility of KNO3\text{KNO}_3 in mol dm−3\text{mol dm}^{-3}.

    Solubility=Moles of KNO3Volume of solution\text{Solubility} = \frac{\text{Moles of KNO}_3}{\text{Volume of solution}} Solubility=0.15446 mol0.060 dm3≈2.57 mol dm−3\text{Solubility} = \frac{0.15446 \text{ mol}}{0.060 \text{ dm}^3} \approx 2.57 \text{ mol dm}^{-3}
    • b) (ii) The solubility of KNO3\text{KNO}_3 is 2.57 mol dm−3\boxed{\text{2.57 mol dm}^{-3}}.

  5. Outline the preparation of PbCl2(s)\text{PbCl}_{2(s)} from Pb(NO3)2\text{Pb(NO}_3)_2 and a mineral acid.

    • c) (i) To prepare PbCl2(s)\text{PbCl}_{2(s)}, mix an aqueous solution of lead(II) nitrate (Pb(NO3)2\text{Pb(NO}_3)_2) with a mineral acid containing chloride ions, such as hydrochloric acid (HCl). Lead(II) chloride will precipitate out as a solid. Filter the precipitate, wash it with distilled water, and then dry it.

  6. Write an equation for the reaction in 2(c)(i).

    • c) (ii) Pb(NO3)2(aq)+2HCl(aq)→PbCl2(s)+2HNO3(aq)\boxed{\text{Pb(NO}_3)_2(aq) + 2\text{HCl}(aq) \to \text{PbCl}_2(s) + 2\text{HNO}_3(aq)}

  7. Give the reason for copper's electron configuration.

    • d) (i) The electron configuration of copper is 1s22s22p63s23p63d104s11s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1 because a completely filled d-subshell (3d103d^{10}) is more stable than a partially filled d-subshell (3d93d^9). One electron from the 4s orbital is promoted to the 3d orbital to achieve this greater stability.

  8. Mention the two oxidation states of iron in its chlorides.

    • d) (ii) The two oxidation states of iron in its chlorides are +2\boxed{\text{+2}} (in FeCl2\text{FeCl}_2) and +3\boxed{\text{+3}} (in FeCl3\text{FeCl}_3).

  9. Name one chemical process in which iron is used as a catalyst.

    • d) (iii) Iron is used as a catalyst in the Haber process\boxed{\text{Haber process}} (for the synthesis of ammonia).
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