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Calculate the gradient of the graph and the enthalpy of solution using the provided data and formulas.

Chemistry

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Answers

2 parts
a
-19.4
b

372 J mol−1372 \, \text{J mol}^{-1}

Step-by-step solution

4 steps

Here are the calculations based on the provided information:

Part a· 2 steps

Calculate the gradient of the graph.

  1. Identify the coordinates of the two points used for the gradient calculation.

    From the student's work, the points are:
    Point 1: (x1,y1)=(0.00327,0.091)(x_1, y_1) = (0.00327, 0.091)
    Point 2: (x2,y2)=(0.00291,0.098)(x_2, y_2) = (0.00291, 0.098)

  2. Calculate the gradient using the formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}.

    m=0.098−0.0910.00291−0.00327m = \frac{0.098 - 0.091}{0.00291 - 0.00327} m=0.007−0.00036m = \frac{0.007}{-0.00036} m=−19.4444...m = -19.4444...

    Rounding to three significant figures, the gradient is −19.4\boxed{-19.4}.

Part b

372 J mol−1372 \, \text{J mol}^{-1}

Calculate the enthalpy of solution using the expression ΔH=−gradient×2.3R\Delta H = -\text{gradient} \times 2.3R, where R=8.31 J mol−1K−1R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1}.

  1. Substitute the calculated gradient and the value of RR into the given expression.

    Using the more precise gradient value:

    ΔH=−(−19.4444...)×2.3×8.31 J mol−1K−1\Delta H = -(-19.4444...) \times 2.3 \times 8.31 \, \text{J mol}^{-1} \text{K}^{-1} ΔH=19.4444...×2.3×8.31 J mol−1\Delta H = 19.4444... \times 2.3 \times 8.31 \, \text{J mol}^{-1} ΔH=19.4444...×19.113 J mol−1\Delta H = 19.4444... \times 19.113 \, \text{J mol}^{-1} ΔH=372.093... J mol−1\Delta H = 372.093... \, \text{J mol}^{-1}
  2. Round the final answer to an appropriate number of significant figures.

    Rounding to three significant figures:

    ΔH=372 J mol−1\Delta H = 372 \, \text{J mol}^{-1}

    The enthalpy of solution is 372 J mol−1\boxed{372 \, \text{J mol}^{-1}}.

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