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What is a standard solution? What mass of sodium carbonate is needed to prepare 500 cm 3 of 0.2 moldm -3 solution? How much water is needed to dilute the solution in a (ii) to 0.04 moldm -3 ?

Chemistry

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Answer

e

Lattice energy: energy released on forming 1 mol ionic solid from gaseous ions

Step-by-step solution

7 steps

1(a)

Molar mass of Na2CO3\mathrm{Na_2CO_3}:
2×23+12+3×16=46+12+48=106 g/mol2\times23 + 12 + 3\times16 = 46 + 12 + 48 = 106\,\mathrm{g/mol}

  1. Step 1

    Mass=n×M\text{Mass} = n \times M
    =0.4 mol×106 g/mol= 0.4\,\mathrm{mol} \times 106\,\mathrm{g/mol}

    =42.4 g= 42.4\,\mathrm{g}

    42.4 g

    1(b)

  2. Step 2

    c=nVc = \frac{n}{V}
    V=ncV = \frac{n}{c}
    =0.4 mol0.3 mol/L= \frac{0.4\,\mathrm{mol}}{0.3\,\mathrm{mol/L}}

    =1.33 L= 1.33\,\mathrm{L}

    1.33 L

    2(a)

  3. Step 3

    n=c×Vn = c \times V
    =0.2 mol/L×0.5 L= 0.2\,\mathrm{mol/L} \times 0.5\,\mathrm{L}

    =0.1 mol= 0.1\,\mathrm{mol}
  4. Step 4

    Mass=n×M\text{Mass} = n \times M
    =0.1 mol×106 g/mol= 0.1\,\mathrm{mol} \times 106\,\mathrm{g/mol}

    =10.6 g= 10.6\,\mathrm{g}

    10.6 g

    2(b)

    Current: 0.5 L0.5\,\mathrm{L} of 0.2 M0.2\,\mathrm{M}, n=0.1 moln=0.1\,\mathrm{mol}

  5. Step 5

    New c=0.1 Mc=0.1\,\mathrm{M}
    V′=nc′=0.1 mol0.1 mol/LV' = \frac{n}{c'} = \frac{0.1\,\mathrm{mol}}{0.1\,\mathrm{mol/L}}

    =1 L= 1\,\mathrm{L}
  6. Step 6

    Water added =V′−V=1 L−0.5 L= V' - V = 1\,\mathrm{L} - 0.5\,\mathrm{L}

    =0.5 L=500 mL= 0.5\,\mathrm{L} = 500\,\mathrm{mL}

    500 mL

    3

    Moles AgNO3\mathrm{AgNO_3}:
    nAg+=0.1 mol/L×0.025 L=0.0025 moln_{\mathrm{Ag^+}} = 0.1\,\mathrm{mol/L} \times 0.025\,\mathrm{L} = 0.0025\,\mathrm{mol}

    Moles NaCl\mathrm{NaCl}:
    nCl−=0.2 mol/L×0.020 L=0.004 moln_{\mathrm{Cl^-}} = 0.2\,\mathrm{mol/L} \times 0.020\,\mathrm{L} = 0.004\,\mathrm{mol}

    NaCl\mathrm{NaCl} solution is in excess (more Cl−\mathrm{Cl^-}).

  7. Step 7

    Excess Cl−=0.004 mol−0.0025 mol\mathrm{Cl^-} = 0.004\,\mathrm{mol} - 0.0025\,\mathrm{mol}

    =0.0015 mol= 0.0015\,\mathrm{mol}

    NaCl solution; 0.0015 mol

    4

    Type of cellCoordination numberNumber of particles per cellSimple cubic61Face-centered cubic124Body-centered cubic82\begin{array}{|c|c|c|} \hline \text{Type of cell} & \text{Coordination number} & \text{Number of particles per cell} \\ \hline \text{Simple cubic} & 6 & 1 \\ \hline \text{Face-centered cubic} & 12 & 4 \\ \hline \text{Body-centered cubic} & 8 & 2 \\ \hline \end{array}

    (d)
    Body-centered cubic structure is used to calculate lattice energy for metals like Fe, Cr, etc.

    (e)
    Lattice energy is the energy released when one mole of an ionic compound in gaseous state forms a solid crystal lattice.
    Lattice energy: energy released on forming 1 mol ionic solid from gaseous ions

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