Indicate how you would prepare each of the following solutions- (a) 1.docx
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28 Jan 2023•0 j'aime•2 vues
Indicate how you would prepare each of the following solutions- (a) 1.docx
28 Jan 2023•0 j'aime•2 vues
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Formation
Indicate how you would prepare each of the following solutions: (a) 12.0 liters of 6.00 M KOH from solid KOH. (b) 100.0 grams of 10.0%(w/w) solution of Na2CO3 from solid Na2CO3. Solution a) Since Molarity is defined as, Molarity = no of moles/ Volume of solution in litres Substituting values, we get, 6 = n/12 Therefore, n = 6*12 = 72 moles = 72*56 g of KOH = 4032g of KOH Therefore given solution can be prepared by adding 4.032kg of KOH in 12 litres of water b) 10% (w/w) solution means 10g of solute per 100g of water Let x grams of Na2CO3 be added, (Weight of Na2CO3/Weight of water) * 100 = 10 x/(100-x) = 0.1 x = 10-0.1x 1.1x = 10 x = 9.09 g Therefore given solution can be prepared by adding 9.09 g of Na2CO3 in 90.90 g of water. .
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Indicate how you would prepare each of the following solutions- (a) 1.docx
- 1. Indicate how you would prepare each of the following solutions: (a) 12.0 liters of 6.00 M KOH from solid KOH. (b) 100.0 grams of 10.0%(w/w) solution of Na2CO3 from solid Na2CO3. Solution a) Since Molarity is defined as, Molarity = no of moles/ Volume of solution in litres Substituting values, we get, 6 = n/12 Therefore, n = 6*12 = 72 moles = 72*56 g of KOH = 4032g of KOH Therefore given solution can be prepared by adding 4.032kg of KOH in 12 litres of water b) 10% (w/w) solution means 10g of solute per 100g of water Let x grams of Na2CO3 be added, (Weight of Na2CO3/Weight of water) * 100 = 10 x/(100-x) = 0.1 x = 10-0.1x 1.1x = 10 x = 9.09 g Therefore given solution can be prepared by adding 9.09 g of Na2CO3 in 90.90 g of water.