1.29 g of a solid carbonate, M2CO3, was dissolved in water and the volume of the solution was made up to 250 cm3 with distilled water in a volumetric flask. 25.0 cm3 of the carbonate solution required 21.9 cm3 of 0.0850 mol dm−3 nitric acid to reach the end point in a titration.
What is the molar mass of the carbonate?
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chem titration question watch
- Thread Starter
- 06-01-2017 15:51
- 06-01-2017 17:26
I think this is how you do it.
1. (21.9/1000) x 0.0850 = 1.8615x10^-3mol of nitric acid
2. This will be the moles of carbonate too as I presume it is a 1:1 ratio in 25cm^3
3. 1.8615x10^-3 x 10 = 0.018615mol of carbonate compound in 250cm^3
4. 1.29/0.018615 = 69.3 (all molar masses to 1dp)
Would it be possible to show the question as it is written?