dissolve 10g of TCA in 100ml distilled water.
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6 litres of 50% + 4 litres of 25%
7 liters of a 20% acid solution consists of 1.4 liters of acid (20% of the total volume) mixed with 5.6 liters of water (80% of the total volume). The amount of acid isn't going to change in the new solution. You are just going to add enough water to make it a 10% solution instead of a 20% solution. So it will be more dilute. That means that 1.4 liters of acid will represent 1/10 of the volume of the new solution. So the total volume of the new solution will be 10 x 1.4 or 14 liters. The amount of water in the new solution will be 14 - 1.4 = 12.6 liters. That is a difference of 12.6 - 5.6 = 7 liters from the amount of water you started with. So you need to add 7 liters of water to the original 20% solution to make it a 10% solution. This makes sense because if you double the amount of the mixture from 7 liters to 14 liters and the amount of acid is unchanged, the solution will be half as strong.
You have 6 litres of alcohol in 24 litres of water You need to add x litres to make 6 equal to 15% of 30 + x. 6 is 15% of 40, so x = 10
25% ammonia solution means 25g Ammonia in 100ml of water. To convert it into moles i.e for 1000ml solvent, 25x100=250 which means 250g of ammonia is present in 1000ml of water to make a 25%solution. To convert it into molar solution we will divide 250 by the molecular mass of ammonia i.e 17. 250/17= 14.71M This means that 25% ammonia solution contains 14.71 moles of ammonia. Now we will use the formula M1V1=M2V2 Let's say we want to make 10 molar solution in 500ml water. Then, M1=14.71 V1=? M2=10 V2=500 Putting values in the above equation: 14.71xV1 = 10x500 V1 = 10x500 / 14.71 = 340ml We will take 340ml of the 25% ammonia solution and make the volume up to 500ml with water. This will be our 10 molar solution of Ammonia.
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