How Do You Calculate the Solubility of Benzoic Acid in Water?

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freedom856
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In an experiment, 7g benzoic acid is dissolving to 350ml distillated water in 500ml conical flask at 60C for a period of time. 100ml of solution is then withdrawn and place at 40C water bath. After several minutes, 10ml sample is withdrawn and the weight of sample is 9.58g. The solution is then transfer to 100ml flask and 20ml water is added. The solution is then titrated against 0.0206M NaOH. Volume of NaOH used is 28.9ml. I wonder how to calculate the solubility of benzoic acid in moles per 1kg water

Can anyone teach me how to calculate this question?

Also, if i have 2 values of the solubility. how can i calculate the average error? Thank you
 
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number of moles of acid react
 
There appears to be some redundant information in this problem. You might object it is put there to confuse you :devil: but in practice in the laboratory you might have it and be confused, you do have to see what is essential.
 
freedom856 said:
number of moles of acid react

So, what is mass of the acid in 9.58g of the solution?
 
Is (0.0206x28.9/1000)x122.12 = 0.0727g ?
 
And you still don't know what to do? What is mass of solvent?
 
o i see. i wonder whether different volume of sample may or may not affect number of moles. Thank you. Is mass of solvent = 9.58-0.0727 = 9.5073g?
 
That's the correct approach. I have not checked numbers.
 
what about average error?