bigbeefbaby
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Question a sample of 2.27g of hydrated barium hydroxide Ba(OH)2.8H2O was dissolved in 1dm3 of water. 25cm3 of this solution was titrated with 0.02 mol dm-3 HCL

Mark Scheme answer:
Moles of Ba(OH)2.8H2O = 2.27/315.3 = 7.20 × 10−3 mol Moles of Ba(OH)2 in 25.0 cm3 = 7.2 x 10-3/40 = 1.80 × 10−4 mol Moles of HCl required = 1.80 × 10−4 × 2
To get moles of Ba(OH)2 why do you have to divide by 40?
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Student 999
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(Original post by bigbeefbaby)
Question a sample of 2.27g of hydrated barium hydroxide Ba(OH)2.8H2O was dissolved in 1dm3 of water. 25cm3 of this solution was titrated with 0.02 mol dm-3 HCL

Mark Scheme answer:
Moles of Ba(OH)2.8H2O = 2.27/315.3 = 7.20 × 10−3 mol Moles of Ba(OH)2 in 25.0 cm3 = 7.2 x 10-3/40 = 1.80 × 10−4 mol Moles of HCl required = 1.80 × 10−4 × 2
To get moles of Ba(OH)2 why do you have to divide by 40?
It’s finding the amount of moles in a 25cm3 solution.So they converted 25cm3 to 1/40dm3.
Then multiplied (7.2 x 10³)(1/40) to find moles.This is the same as dividing by 40
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Pigster
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(Original post by bigbeefbaby)
Question a sample of 2.27g of hydrated barium hydroxide Ba(OH)2.8H2O was dissolved in 1dm3 of water. 25cm3 of this solution was titrated with 0.02 mol dm-3 HCL

Mark Scheme answer:
Moles of Ba(OH)2.8H2O = 2.27/315.3 = 7.20 × 10−3 mol Moles of Ba(OH)2 in 25.0 cm3 = 7.2 x 10-3/40 = 1.80 × 10−4 mol Moles of HCl required = 1.80 × 10−4 × 2
To get moles of Ba(OH)2 why do you have to divide by 40?
The shorter answer is, the MS is designed primarily to help the markers mark, helping students in the future is secondary to the MS doing its primary job.
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deskochan
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Moles of Ba(OH)2.8H2O = 2.27/315.3 = 7.20 × 10−3 mol
Thus, the concentration of Ba(OH)2 in 1 dm3 = 7.2 x 10-3 moldm-3
Moles of Ba(OH)2 in 25.0 cm3 = 7.2 x 10-3 x 25 cm3 = 7.2 x 10-3 x (25/1000) = 7.2 x 10-3 x (1/40)
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