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# Science: If I have 8.2 grams of sugar in 25 ml of water for

my initial solution and transfer...
Science: If I have 8.2 grams of sugar in 25 ml of water for my initial solution and transfer 2.5 ml out and then refill the solution to 25 ml what will the final concentration in Moles be? Then if again I have my new solution and now trnasfer 4.5 ml out and then refill the remasining solution up to 25 ml of water, what is my final concentration in moles?
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9/8/2012
Ryan, Engineer
Category: Homework
Satisfied Customers: 9,096
Experience: B.S. in Civil Engineering
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Hi,

Welcome! Thank you for using JustAnswer.

Assuming that the sugar is common table sugar (sucrose), the molecular weight is 342.30 g/mol.

The concentration of the original solution is:

(8.2 g)/ [ (25 ml)(1 L/1000ml)(342.30 g/mol) ] = 0.9582 M

Replacing 2.5 ml of this solution with water would make the new concentration:

(0.9582 M) (25 ml - 2.5 ml) / (25 ml) = 0.8624 M

Then, replacing 4.5 ml of this new solution with water would make the second new concentration:

(0.8624 M)(25 ml - 4.5 ml) / (25 ml) = 0.7072 M

Note that the concentrations here are in moles per liter (designated with the letter M).

If you want the amount of sugar in each solution, in moles, that would be:

Original solution: (0.9582 M)(25 ml / 1000 ml/L) = 0.0240 moles

First new solution: (0.8624 M)(25 ml / 1000 ml/L) = 0.0216 moles

Second new solution: (0.7072 M)(25 ml / 1000 ml/L) = 0.0177 moles

Thanks,

Ryan
Customer reply replied 5 years ago

Does the first solution remove any soution. Second should remove 2.5, third should remove 4,5 ml, then 3ml, then 6ml. I need the final concentration of each after each amount is transferred out in (M)

Hi,

The first value, 0.9582 M, was for the original mix, before any of the solution is removed.

The second value, 0.8624 M, was for the solution after removing 2.5 ml of the first solution.

The third value, 0.7072 M, was for the solution after removing 4.5 ml of the second solution.

The parts about removing 3 ml and 6 ml were not in my previous solution because that information was not included in your original post. Here are the answers for those steps:

Starting with the third solution (0.7072 M), removing 3 ml and then replacing it with water leaves a concentration of:

(0.7072 M)(25 ml - 3 ml) / 25 ml = 0.6223 M

Removing 6 ml of this solution and replacing it with water leaves a concentration of:

(0.6223 M)(25 ml - 6 ml)/ 25 ml = 0.4729 M

In table form, the solutions are:

original solution: 0.9582 M
after replacing 2.5 ml: 0.8624 M
after replacing 4.5 ml: 0.7072 M
after replacing 3 ml: 0.6223 M
after replacing 6 ml: 0.4729 M

Thanks,

Ryan
Ryan, Engineer
Category: Homework
Satisfied Customers: 9,096
Experience: B.S. in Civil Engineering
Verified
Customer reply replied 5 years ago

So can I make the assumption that as the sugar shrinks from the solution the weight in grams rises but the molecular weight in moles decreases? Is that right?

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Ryan, Engineer
Category: Homework
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