A stopwatch was started when 50 cm³ of a 0.20 M sodium thiosulfate solution was poured into a conical flask containing 10 cm³ of 1.0 M HCl solution - Leaving Cert Chemistry - Question 3 - 2010

To prepare 50 cm³ of a 0.12 M sodium thiosulfate solution from a 0.20 M stock solution, the following steps should be taken:

1. Calculate the volume of 0.20 M solution needed using the dilution formula:

   $C_1V_1 = C_2V_2$

   where $C_1 = 0.20$ M, $C_2 = 0.12$ M, and $V_2 = 50$ cm³:

   $V_1 = \frac{C_2V_2}{C_1} = \frac{0.12 \times 50}{0.20} = 30\text{ cm}^3$

2. Measure 30 cm³ of the 0.20 M sodium thiosulfate solution.

3. Add distilled or deionized water to the 30 cm³ solution to make up the final volume to 50 cm³.

4. Mix the solution thoroughly.

To plot the graph, use the provided data to create a scatter plot of the concentration (M) versus the reaction rate (1/t, min⁻¹). Ensure that the x-axis is labeled as 'Concentration (M)' and the y-axis as 'Rate (1/t min⁻¹)'. All points should be plotted accurately, and a straight line should connect them to show the trend.

The graph demonstrates a direct proportional relationship between the concentration of sodium thiosulfate and the reaction rate, indicating that as the concentration of sodium thiosulfate increases, the rate of reaction also increases.

To predict the time for the reaction to occur, one can extrapolate from the graph based on the concentration. For 0.20 M sodium thiosulfate, the time taken was recorded at 1.4 minutes, hence if the temperature increases, one could anticipate a shorter reaction time.

The time taken would be expected to decrease as the temperature increases. This is because increasing the temperature generally raises the kinetic energy of the reactant particles, leading to more frequent and effective collisions, thus increasing the reaction rate.