Solve for me from 4-7

Mathematics

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Here are the solutions to questions 4-7:

4. A solution of NaOH\text{NaOH} contains 4.0g of NaOH\text{NaOH} in 250ml of solution. Calculate the molarity of the solution. (Na=23\text{Na}=23, O=16\text{O}=16, H=1\text{H}=1)

  1. Calculate the molar mass of NaOH\text{NaOH}.

    Molar mass of NaOH=(1×23)+(1×16)+(1×1)=23+16+1=40 g/mol\text{Molar mass of NaOH} = (1 \times 23) + (1 \times 16) + (1 \times 1) = 23 + 16 + 1 = 40 \text{ g/mol}

  2. Calculate the number of moles of NaOH\text{NaOH}.

    Moles of NaOH=MassMolar mass=4.0 g40 g/mol=0.10 mol\text{Moles of NaOH} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{4.0 \text{ g}}{40 \text{ g/mol}} = 0.10 \text{ mol}

  3. Convert the volume of the solution from milliliters to liters.

    Volume of solution=250 mL=0.250 L\text{Volume of solution} = 250 \text{ mL} = 0.250 \text{ L}

  4. Calculate the molarity of the solution.

    Molarity=Moles of soluteVolume of solution (L)=0.10 mol0.250 L=0.4 M\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}} = \frac{0.10 \text{ mol}}{0.250 \text{ L}} = \boxed{\text{0.4 M}}

    5. Calculate the mass of Na2CO3\text{Na}_2\text{CO}_3 required to prepare 250ml of 0.1M solution. (Na=23\text{Na}=23, C=12\text{C}=12, O=16\text{O}=16)

  5. Calculate the molar mass of Na2CO3\text{Na}_2\text{CO}_3.

    Molar mass of Na2CO3=(2×23)+(1×12)+(3×16)=46+12+48=106 g/mol\text{Molar mass of Na}_2\text{CO}_3 = (2 \times 23) + (1 \times 12) + (3 \times 16) = 46 + 12 + 48 = 106 \text{ g/mol}

  6. Convert the volume of the solution from milliliters to liters.

    Volume of solution=250 mL=0.250 L\text{Volume of solution} = 250 \text{ mL} = 0.250 \text{ L}

  7. Calculate the number of moles of Na2CO3\text{Na}_2\text{CO}_3 required.

    Moles of Na2CO3=Molarity×Volume (L)=0.1 M×0.250 L=0.025 mol\text{Moles of Na}_2\text{CO}_3 = \text{Molarity} \times \text{Volume (L)} = 0.1 \text{ M} \times 0.250 \text{ L} = 0.025 \text{ mol}

  8. Calculate the mass of Na2CO3\text{Na}_2\text{CO}_3 required.

    Mass of Na2CO3=Moles×Molar mass=0.025 mol×106 g/mol=2.65 g\text{Mass of Na}_2\text{CO}_3 = \text{Moles} \times \text{Molar mass} = 0.025 \text{ mol} \times 106 \text{ g/mol} = \boxed{\text{2.65 g}}

    6. Calculate the volume of 0.2M HCl\text{HCl} required to neutralize 25ml of 0.1M NaOH\text{NaOH}. (HCl+NaOH→NaCl+H2O\text{HCl} + \text{NaOH} \to \text{NaCl} + \text{H}_2\text{O})

  9. Write the balanced chemical equation and determine the mole ratio.

    The given equation is HCl+NaOH→NaCl+H2O\text{HCl} + \text{NaOH} \to \text{NaCl} + \text{H}_2\text{O}.
    From the equation, 1 mole of HCl\text{HCl} reacts with 1 mole of NaOH\text{NaOH}. The mole ratio is 1:11:1.

  10. Convert the volume of NaOH\text{NaOH} solution from milliliters to liters.

    Volume of NaOH=25 mL=0.025 L\text{Volume of NaOH} = 25 \text{ mL} = 0.025 \text{ L}

  11. Calculate the number of moles of NaOH\text{NaOH}.

    Moles of NaOH=Molarity×Volume (L)=0.1 M×0.025 L=0.0025 mol\text{Moles of NaOH} = \text{Molarity} \times \text{Volume (L)} = 0.1 \text{ M} \times 0.025 \text{ L} = 0.0025 \text{ mol}

  12. Use the mole ratio to find the moles of HCl\text{HCl} required.

    Since the mole ratio of HCl\text{HCl} to NaOH\text{NaOH} is 1:11:1,

    Moles of HCl=Moles of NaOH=0.0025 mol\text{Moles of HCl} = \text{Moles of NaOH} = 0.0025 \text{ mol}

  13. Calculate the volume of HCl\text{HCl} solution required.

    Volume of HCl (L)=Moles of HClMolarity of HCl=0.0025 mol0.2 M=0.0125 L\text{Volume of HCl (L)} = \frac{\text{Moles of HCl}}{\text{Molarity of HCl}} = \frac{0.0025 \text{ mol}}{0.2 \text{ M}} = 0.0125 \text{ L}

    Convert to milliliters:

    Volume of HCl=0.0125 L×1000 mL/L=12.5 mL\text{Volume of HCl} = 0.0125 \text{ L} \times 1000 \text{ mL/L} = \boxed{\text{12.5 mL}}

    7. Calculate the percentage by mass of water of crystallization in Na2CO3⋅10H2O\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}. (Na=23\text{Na}=23, C=12\text{C}=12, O=16\text{O}=16, H=1\text{H}=1)

  14. Calculate the molar mass of Na2CO3⋅10H2O\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}.

    Molar mass of Na2CO3=(2×23)+(1×12)+(3×16)=46+12+48=106 g/mol\text{Molar mass of Na}_2\text{CO}_3 = (2 \times 23) + (1 \times 12) + (3 \times 16) = 46 + 12 + 48 = 106 \text{ g/mol} Molar mass of H2O=(2×1)+(1×16)=2+16=18 g/mol\text{Molar mass of H}_2\text{O} = (2 \times 1) + (1 \times 16) = 2 + 16 = 18 \text{ g/mol} Molar mass of 10H2O=10×18=180 g/mol\text{Molar mass of 10H}_2\text{O} = 10 \times 18 = 180 \text{ g/mol} Total molar mass of Na2CO3⋅10H2O=106+180=286 g/mol\text{Total molar mass of Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} = 106 + 180 = 286 \text{ g/mol}

  15. Calculate the percentage by mass of water of crystallization.

    Percentage by mass of H2O=Mass of 10H2OTotal molar mass of Na2CO3⋅10H2O×100%\text{Percentage by mass of H}_2\text{O} = \frac{\text{Mass of 10H}_2\text{O}}{\text{Total molar mass of Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}} \times 100\% =180 g/mol286 g/mol×100%≈62.94%= \frac{180 \text{ g/mol}}{286 \text{ g/mol}} \times 100\% \approx \boxed{\text{62.94\%}}
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