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Draw and label the axes. The x-axis should represent Temperature (^ C) and the y-axis should represent Solubility (g/100g of water).

Mathematics

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2 parts
b(i)

130 g/100g of water\text{130 g/100g of water}

b(ii)

75.4 g\text{75.4 g}

Step-by-step solution

9 steps

Part a· 4 steps

To plot the graph of solubility of potassium nitrate against temperature

  1. Draw and label the axes.

    The x-axis should represent Temperature (∘C^\circ \text{C}) and the y-axis should represent Solubility (g/100g of water).

  2. Choose an appropriate scale for each axis.

    • For the x-axis (Temperature): A suitable scale would be 1 cm1 \text{ cm} (or 1 major grid square) representing 10∘C10^\circ \text{C}. This allows the temperature range from 10∘C10^\circ \text{C} to 100∘C100^\circ \text{C} to fit well.
    • For the y-axis (Solubility): A suitable scale would be 1 cm1 \text{ cm} (or 1 major grid square) representing 10 g/100g of water10 \text{ g/100g of water}. Start the y-axis at a value like 70 g/100g of water70 \text{ g/100g of water} to accommodate the lowest solubility value of 80 g/100g of water80 \text{ g/100g of water}.
  3. Plot the given data points from the table onto the graph paper.

    For example, plot $(10, 80)$, $(20, 88)$, $(30, 96)$, and so on, up to $(100, 180)$.

  4. Draw a smooth curve through the plotted points.

    This curve represents the solubility curve for potassium nitrate.

Part b(i)

130 g/100g of water\text{130 g/100g of water}

The solubility of potassium nitrate at 65∘C65^\circ \text{C}.

From the graph, determine

(Note: Since a physical graph cannot be plotted here, the values are determined by interpolation from the given data, which is equivalent to reading from a well-drawn graph.)

  1. Locate 65∘C65^\circ \text{C} on the temperature (x-axis).

  2. Draw a vertical line from 65∘C65^\circ \text{C} up to the solubility curve.

  3. From the point where the vertical line intersects the curve, draw a horizontal line to the solubility (y-axis).

    Based on interpolation between 60∘C60^\circ \text{C} (124 g/100g water) and 70∘C70^\circ \text{C} (136 g/100g water):

    Solubility at 65∘C=124+65−6070−60×(136−124)\text{Solubility at } 65^\circ \text{C} = 124 + \frac{65-60}{70-60} \times (136-124) =124+510×12= 124 + \frac{5}{10} \times 12 =124+0.5×12= 124 + 0.5 \times 12 =124+6=130 g/100g of water= 124 + 6 = 130 \text{ g/100g of water}

    The solubility of potassium nitrate at 65∘C65^\circ \text{C} is 130 g/100g of water\boxed{\text{130 g/100g of water}}.

Part b(ii)

75.4 g\text{75.4 g}

The mass of potassium nitrate that would dissolve in 50g50 \text{g} of water at 82∘C82^\circ \text{C}.

  1. Determine the solubility at 82∘C82^\circ \text{C} from the graph.

    Based on interpolation between 80∘C80^\circ \text{C} (148 g/100g water) and 90∘C90^\circ \text{C} (162 g/100g water):

    Solubility at 82∘C=148+82−8090−80×(162−148)\text{Solubility at } 82^\circ \text{C} = 148 + \frac{82-80}{90-80} \times (162-148) =148+210×14= 148 + \frac{2}{10} \times 14 =148+0.2×14= 148 + 0.2 \times 14 =148+2.8=150.8 g/100g of water= 148 + 2.8 = 150.8 \text{ g/100g of water}

    This means 150.8 g150.8 \text{ g} of potassium nitrate dissolves in 100 g100 \text{ g} of water at 82∘C82^\circ \text{C}.

  2. Calculate the mass that dissolves in 50 g50 \text{ g} of water.

    Since 50 g50 \text{ g} of water is half of 100 g100 \text{ g} of water, the mass of potassium nitrate that dissolves will also be half.

    Mass of KNO3=150.8 g100 g water×50 g water\text{Mass of KNO}_3 = \frac{150.8 \text{ g}}{100 \text{ g water}} \times 50 \text{ g water} =150.82 g=75.4 g= \frac{150.8}{2} \text{ g} = 75.4 \text{ g}

    The mass of potassium nitrate that would dissolve in 50g50 \text{g} of water at 82∘C82^\circ \text{C} is 75.4 g\boxed{\text{75.4 g}}.

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