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.
