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Question Q3i) Give the characteristic E.C of group VII element ii) How does electronic affinity of group VII element down the group

Chemistry

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Part a· 4 steps

ii) Calculate the molality of the solution containing 2.50g of Na oxalate C2O4 in 250 cm³ of solution.

To calculate molality, we need the mass of the solvent in kilograms. The problem provides the volume of the solution (250 cm³). Without the density of the solution, we cannot determine the mass of the solvent. Therefore, molality cannot be calculated with the given information.

However, if the question intended to ask for molarity (moles of solute per liter of solution), here is the calculation:

  1. Calculate the molar mass of sodium oxalate (Na2C2O4\text{Na}_2\text{C}_2\text{O}_4).

    Molar mass of Na=22.99 g/mol\text{Na} = 22.99 \text{ g/mol}
    Molar mass of C=12.01 g/mol\text{C} = 12.01 \text{ g/mol}
    Molar mass of O=16.00 g/mol\text{O} = 16.00 \text{ g/mol}

    M(Na2C2O4)=(2×22.99)+(2×12.01)+(4×16.00)M(\text{Na}_2\text{C}_2\text{O}_4) = (2 \times 22.99) + (2 \times 12.01) + (4 \times 16.00) M(Na2C2O4)=45.98+24.02+64.00=134.00 g/molM(\text{Na}_2\text{C}_2\text{O}_4) = 45.98 + 24.02 + 64.00 = 134.00 \text{ g/mol}

  2. Calculate the number of moles of Na2C2O4\text{Na}_2\text{C}_2\text{O}_4.

    Moles=MassMolar mass=2.50 g134.00 g/mol≈0.01866 mol\text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{2.50 \text{ g}}{134.00 \text{ g/mol}} \approx 0.01866 \text{ mol}

  3. Convert the volume of the solution to dm3\text{dm}^3 (liters).

    Volume=250 cm3=250×10−3 dm3=0.250 dm3\text{Volume} = 250 \text{ cm}^3 = 250 \times 10^{-3} \text{ dm}^3 = 0.250 \text{ dm}^3

  4. Calculate the molarity.

    Molarity=Moles of soluteVolume of solution (dm3)=0.01866 mol0.250 dm3\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution (dm}^3)} = \frac{0.01866 \text{ mol}}{0.250 \text{ dm}^3} Molarity≈0.0746 mol dm−3\text{Molarity} \approx 0.0746 \text{ mol dm}^{-3}

    The molarity of the solution is 0.0746 mol dm−3\boxed{0.0746 \text{ mol dm}^{-3}}.

Part b· 3 steps

0.13g of a liquid was evaporated in a 500cm³ flask at 25°C. If the pressure of the liquid was found to be 152mmHg, what is the relative molecular mass of the liquid?

  1. Convert the given values to appropriate units for the ideal gas law.

    Mass (mm) = 0.13 g0.13 \text{ g}
    Volume (VV) = 500 cm3=0.500 L500 \text{ cm}^3 = 0.500 \text{ L}
    Temperature (TT) = 25∘C=25+273.15=298.15 K25^\circ\text{C} = 25 + 273.15 = 298.15 \text{ K}
    Pressure (PP) = 152 mmHg152 \text{ mmHg}
    Convert pressure from mmHg to atmospheres (atm):

    P=152 mmHg×1 atm760 mmHg=0.20 atmP = 152 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}} = 0.20 \text{ atm}

    The ideal gas constant (RR) is 0.0821 L atm mol−1 K−10.0821 \text{ L atm mol}^{-1} \text{ K}^{-1}.

  2. Use the ideal gas law to find the number of moles (nn).

    The ideal gas law is PV=nRTPV = nRT.
    Rearranging for nn:

    n=PVRT=(0.20 atm)×(0.500 L)(0.0821 L atm mol−1 K−1)×(298.15 K)n = \frac{PV}{RT} = \frac{(0.20 \text{ atm}) \times (0.500 \text{ L})}{(0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1}) \times (298.15 \text{ K})} n=0.1024.47≈0.004086 moln = \frac{0.10}{24.47} \approx 0.004086 \text{ mol}

  3. Calculate the relative molecular mass (MrM_r).

    The number of moles (nn) is also equal to massmolar mass\frac{\text{mass}}{\text{molar mass}}.

    n=mMr  ⟹  Mr=mnn = \frac{m}{M_r} \implies M_r = \frac{m}{n} Mr=0.13 g0.004086 mol≈31.81 g/molM_r = \frac{0.13 \text{ g}}{0.004086 \text{ mol}} \approx 31.81 \text{ g/mol}

    Rounding to two significant figures (due to 0.13 g and 0.20 atm):
    The relative molecular mass of the liquid is 32 g/mol\boxed{32 \text{ g/mol}}.

Part c

Indium has a half life of 12.23 years.

This is a statement of fact and does not contain a question to be answered.

Part g

i) How does electronic affinity of group VII element down the group?

Electronic affinity generally decreases down Group VII. This trend occurs because as you move down the group, the atomic radius increases, meaning the outermost electrons are further from the nucleus. The increased shielding effect from inner electron shells also reduces the effective nuclear charge experienced by an incoming electron. Both factors lead to a weaker attraction for an additional electron, thus decreasing the energy released (or increasing the energy required) when an electron is added to the atom.

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