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Calculate the average volume of solution W used and its concentration, given it's prepared by diluting 75 cm³ of solution K (1 M H₂SO₄) to 250 cm³.

Chemistry

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Answer

a(ii)

0.300 mol/dm3\text{0.300 mol/dm}^3

Step-by-step solution

3 steps

Here are the calculations for the given problem.

Part a(i)

Average volume of solution W used

Calculate the

The table for burette readings (Table 1) is empty. To calculate the average volume of solution W used, experimental data from the titration (final and initial burette readings) is required. Since this data is not provided, the average volume cannot be calculated.

Part a(ii)

0.300 mol/dm3\text{0.300 mol/dm}^3

Concentration of solution W

Solution W is prepared by diluting 75 cm375 \text{ cm}^3 of solution K (1 M H2SO41 \text{ M H}_2\text{SO}_4) to 250 cm3250 \text{ cm}^3.

  1. Convert the volumes to dm3\text{dm}^3.

    Initial volume of solution K (VKV_K) = 75 cm3=751000 dm3=0.075 dm375 \text{ cm}^3 = \frac{75}{1000} \text{ dm}^3 = 0.075 \text{ dm}^3
    Final volume of solution W (VWV_W) = 250 cm3=2501000 dm3=0.250 dm3250 \text{ cm}^3 = \frac{250}{1000} \text{ dm}^3 = 0.250 \text{ dm}^3

  2. Use the dilution formula C1V1=C2V2C_1V_1 = C_2V_2 to find the concentration of solution W (CWC_W).

    Here, CKVK=CWVWC_K V_K = C_W V_W.

    (1 mol/dm3)×(0.075 dm3)=CW×(0.250 dm3)(1 \text{ mol/dm}^3) \times (0.075 \text{ dm}^3) = C_W \times (0.250 \text{ dm}^3)

  3. Solve for CWC_W.

    CW=1 mol/dm3×0.075 dm30.250 dm3C_W = \frac{1 \text{ mol/dm}^3 \times 0.075 \text{ dm}^3}{0.250 \text{ dm}^3} CW=0.0750.250 mol/dm3C_W = \frac{0.075}{0.250} \text{ mol/dm}^3 CW=0.300 mol/dm3C_W = \boxed{\text{0.300 mol/dm}^3}
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