See your other post. Note that C1V1 = C2V2 is the same as Ma x mLa = Mb x mLb
and note that 8.44 x 14.8 = 250 x 0.500 give or take a little for rounding
You are a work study student in our chemistry department. Amy, your supervisor has just asked you to prepare 250ml of 0.500M NH3 solution for tomorrow's undergraduate experiment. The Stock Solutions cabinet is under the Stockroom Explorer. You will find find a 2.50 liter bottle containing 14.8M NH3. Please prepare a flask containing 250ml of a 0.500M solution. To ensure proper credit, please write your calculations in the space given below.
Need some help?
In designing your procedure, it is useful to remember that diluting a substance does not change the amount of that substance. So the number of moles of the substance before the dilution is the same as the number of moles after the dilution (conservation of matter). The number of moles in solution is given by the concentration (C in moles/liter) multiplied by the volume (V in liters). So the conservation of matter implies:
C1V1 = C2V2
where C is concentration and V is volume. 1 labels the solution before the dilution and 2 labels the solution after the dilution. (This expression is based on the number of moles of the substance between the same before and after the dilution, with C (moles/liter) * V (liter) = moles).
Important: Please describe your complete procedure and the key quantities you measure. Points are based on whether or not you explain your procedure in sufficient detail for us to know what you did. You are not graded on the method you used; all approaches that produce accurate results are fine.
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