Asked by Basil salvation
25cm3 of 0.1 mole H2SO4 neutralizes 20.0cm3 of impure NaOH containing 3.0g of impure NaOH in 250cm3 Of a solution . calculate the percentage purity of the NaOH (Na=23,O=16,H=1)
Answers
Answered by
DrBob222
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
mols H2SO4 = M x L 0.1M x 0.025 L = 0.0025 mols.
mols NaOH neutralized is 2 x that = 0.0050
That's the amount of pure NaOH in the 20 cc. So how much is in the 250 cc container? That will be 0.005 x (250/20) = ? mols NaOH
grams NaOH in the 3.0 g sample is ?mols NaOH from above x 40 g/mol = ?
The %purity = (grams pure NaOH/3.0)*100 = ?
I get something close to 80% purity.
mols H2SO4 = M x L 0.1M x 0.025 L = 0.0025 mols.
mols NaOH neutralized is 2 x that = 0.0050
That's the amount of pure NaOH in the 20 cc. So how much is in the 250 cc container? That will be 0.005 x (250/20) = ? mols NaOH
grams NaOH in the 3.0 g sample is ?mols NaOH from above x 40 g/mol = ?
The %purity = (grams pure NaOH/3.0)*100 = ?
I get something close to 80% purity.
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