Edexcel Core Practical 1 A-Level Chemistry Urgent Help needed 😳

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Report Thread starter 1 month ago
Hello, whilst revising questions relating to the practical tasks I came across a problem concerning the experiment to measure the molar
volume of a gas, here being carbon dioxide. I am really struggling with these questions and think that I have overcomplicated in my mind.

It is clear that each mole of sodium hydrogen carbonate produces one mole of carbon dioxide. It is stated that in 100 cm^3 the sodium hydrogen carbonate solution has 0.0151 mol.

1) Using the results presented in the table (see attached); calculate the number of moles of sodium hydrogen carbonate used for each volume of standard solution.

Well, I know that the formula relating the number of moles, volume of solution and molarity as number of moles (n)=molarity (M)*volume of solution (V).
Molarity = number of moles/litres of solution
Convert to litres; 100cm^3=0.1 L
Molarity = 0.0151/0.1
Molarity = 0.151 mol/L
Then, using the first formula: number of moles (n)=molarity (M)*volume of solution (V) calculate the number of moles for each volume:
n= 0.151 * 1 = 0.0151 mol
n=0.151 *2.5 = 0.3375 mol
n=0.151 *5 =0.755 mol
n=0.151 *7.5 = 1.1325 mol
n=0.151 *10 =1.15 mol

I really am struggling here and know that I have made a mistake. I am very confused and feel that I am grasping at straws. Have I used the incorrect formula?
Should I have instead considered that since there are 0.0151 mol in 100 cm^3 of the sodium hydrogen carbonate solution this implies that in 1 cm^3 there are 1.51 *10^-4 mol.
Hence, at each volume:
1.0 cm^3 * 1.51 *10^-4 mol = 1.51 *10^-4 mol
2.5 cm^3 * 1.51 *10^-4 mol = 3.775 * 10^-4 mol
5.0 cm^3 * 1.51 *10^-4 mol = 7.55 *10^-4 mol
7.5 cm^3 * 1.51 *10^-4 mol = 1.1325 *10^-3mol
10 cm^3 * 1.51 *10^-4 mol = 1.51 *10^-3 mol
I would appreciate any clarity.

2) Using your results, plot a graph to show the volume of gas produced by the number of moles of sodium hydrogen carbonate.

So, I think that I need to plot a graph of volume of gas on the y-axis by the number of moles (which I have calculated) on the x-axis. However, since the vertical scale has an approximate range of from 0 to 2 x 10^-3 this is rather a horrid scale to work with. Do you think I need to plot a the graph using logarithmic scale instead? I have tried to do this on desmos but I think I have done so correctly. Should I have applied logarithmic scales to both horizontal and vertical axes for the data? I have also attached a graph of this but I am not confident in this either.

3)Using the graph you have drawn, show you working to give the volume of 1 mol of carbon dioxide.

Well, I think that using the formula molar volume=volume of gas / number of moles one can see that this would correspond to the gradient of the graph. Then, in order to find the volume of 1 mol of CO2 would they rearrange this as; volume = molar volume* number of moles
volume = 24000 cm^3*1 mol = 24000 cm^3

I do not think that this is correct because I have not used data from the graph as required. I would appreciate any guidance here also.

Thank you to anyone who replies 👍
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Last edited by LukeWatson4590; 1 month ago

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