Asked by neece
a sucrose solution with a volume of B uL has a density of C g/ml. the solution is D % sucrose by mass. if i added x ml of this sucrose to another solution, how many grams sucrose did I add?a sucrose so
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Answered by
DrBob222
One can get very confused with all of the symbols; what I do is make up a problem, then substitute symbols.
Let's say we have a 10% sucrose soln that has a density of 1.1 g/mL (that may not be right but it doesn't matter since we're just doing the exercise to keep symbols straight) and we want to add 5 mL of it to a beaker. How much sucrose did we add?
10%soln means 10g/100 g soln.
If the density is 1.1 g/mL then 100 g soln has a volume of 100/1.1 = 90.9 mL. If we add 5 mL of that soln then we've added 10 g x (5 mL/90.9 mL) = 0.55 g sucrose. Done in one step that is
10 x 1.1 x (5/100) = 0.55 g sucrose.
So that is D%*density*x mL/100 = g sucrose added to the other beaker. I don't see that the B uL has anything to do with it--that's the amount of solution you have. Obviously B uL must be greater than the x mL that is being added; otherwise the x mL can't be added.
Let's say we have a 10% sucrose soln that has a density of 1.1 g/mL (that may not be right but it doesn't matter since we're just doing the exercise to keep symbols straight) and we want to add 5 mL of it to a beaker. How much sucrose did we add?
10%soln means 10g/100 g soln.
If the density is 1.1 g/mL then 100 g soln has a volume of 100/1.1 = 90.9 mL. If we add 5 mL of that soln then we've added 10 g x (5 mL/90.9 mL) = 0.55 g sucrose. Done in one step that is
10 x 1.1 x (5/100) = 0.55 g sucrose.
So that is D%*density*x mL/100 = g sucrose added to the other beaker. I don't see that the B uL has anything to do with it--that's the amount of solution you have. Obviously B uL must be greater than the x mL that is being added; otherwise the x mL can't be added.
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