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Chemistry A-level moles
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Yash12345
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24.2cm^3 of 0.2M HCl was neutralised by 25cm^3 of an alkaline solution. The alkaline solution contained a 16.47g mixture of sodium and rubidium hydroxide dissolved to give 1000cm^3 of solution. Calculate the mass of NaOh in the mixture.
I have caluclated the moles of HCl (0.0242 x 0.2) but I don't know what to do next.
Any help appreciated.
I have caluclated the moles of HCl (0.0242 x 0.2) but I don't know what to do next.
Any help appreciated.
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charco
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(Original post by Yash12345)
24.2cm^3 of 0.2M HCl was neutralised by 25cm^3 of an alkaline solution. The alkaline solution contained a 16.47g mixture of sodium and rubidium hydroxide dissolved to give 1000cm^3 of solution. Calculate the mass of NaOh in the mixture.
I have caluclated the moles of HCl (0.0242 x 0.2) but I don't know what to do next.
Any help appreciated.
24.2cm^3 of 0.2M HCl was neutralised by 25cm^3 of an alkaline solution. The alkaline solution contained a 16.47g mixture of sodium and rubidium hydroxide dissolved to give 1000cm^3 of solution. Calculate the mass of NaOh in the mixture.
I have caluclated the moles of HCl (0.0242 x 0.2) but I don't know what to do next.
Any help appreciated.
So you have the moles of acid = 4.84 x 10-3
And you know that both of the bases are strong and provide hydroxide ions.
The total moles of hydroxide ions must equal the acid ions = 4.84 x 10-3
If you let the moles of sodium hydroxide = x, then the moles of rubidium hydroxide = total moles of hydroxide - x
You also know the mass of the mixture in 25 ml = 16.47/40 = 0.4118g
if the moles of NaOH = x then the mass due to NaOH = 40x and the mass due to RbOH = (total moles - x)(85.56 + 17)
Hence:
0.4118 = 40x + (total moles - x)(85.56 + 17)
can you take it from here?
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