Answer:
3.29 grams NaCl
Step-by-step explanation:
I assume 0.45m is 0.45 molar. The m is usually capitalized, to prevent confusion with meters, moles, etc. : 0.45M
It is very useful to use of the definition of molar in the thought process.
The definition of molar, M, is moles solute per liter of solution. The term 0.45M NaCl means that a solution has 0.45 moles of NaCl in 1 liter of solution. a half-liter would have 0.225 moles of NaCl, which is equivalent to 0.45 moles/liter. in terms of concentration.
We'll need the molar mass of NaCl, which is:
Na: 23.0
Cl: 35.5
58.5 grams/mole
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Now let's calculate how many moles of NaCl are needed. We want 125ml (0.125L) of 0.45M NaCl.
(0.45 moles NaCl/Liter)
multiply that by the actual volume required:
(0.45 moles NaCl/Liter)*(0.125Liter) = 0.0563 moles NaCl
[The liters unit cancels, leaving only moles NaCl. The units are helpful in understanding the mathematical steps required and help insure a correct answer.]
To get grams NaCl, multiply by the molar mass of NaCl:
(0.0563 moles NaCl)*(58.5 grams NaCl/mole NaCl) = 3.29 grams NaCl
Note the units: moles NaCl cancel, leaving grams NaCl.
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There is a useful relationship to remember concerning the dilution of a solution. It is not needed here, but it does highlight the usefulness of understanding units. When diluting a volume, V1, of a solution of concentration M1 to achieve a concentration and volume of of M2 and V2, we can use:
M1V1 = M2V2
Example: 125 ml of 1M HCl is needed for an experiment. The only supply is 5M HCl. How much of the 5M HCl is needed to make 125ml of the 1M HCl?
M1 = 5M HCl
V1 = The unknown
M2 = 2.5M HCl
V2 = 100 ml
M1V1 = M2V2
(5M)*(V1) = (1M)*(125ml)
V1 = (1M)*(125ml)/(5M)
V1 = 25 ml
25 ml of 5M HCl may be added to 75 ml of water (slowly) to form 100ml of 2.5M HCl.
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Extra material: This is not relevant to the original problem, but it demonstrates the importance of units and the fact that it is the moles of a substance that is the important element.
If the concentration and volume of M1V1 are in moles/liter and liter, the term M1V1 has units of:
(moles/Liter)*(L), which reduces to moles of the substance.
The same with the term on the right, M2V2: it also reduces to moles. So the equation is simply saying that whatever number of moles one starts with, the same number of moles will wind up in the final solution.
A bonus discovery is that the volume units cancel in this calculation. It seems unlikely, but this means the volumes don't need to be adjusted from ml to liters, since volume unit cancels out.
If the original problems stated that the goal is to make 125 ml of 0.45M NaCl from a stock of 1.0M NaCl solution, the calculation would be:
M1 = 1.0M NaCl
V1 = ?
M2 = 0.45M NaCl
V2 = 125 ml
M1V1 = M2V2
(1.0M)*(V1) = (0.45M)*(125ml)
V1 = 56.25 ml
68.75 ml of water may be added to 56.25 ml of NaCl to form 125 ml of 0.45M NaCl.
Although M is expressed as moles/liter, we don't need to convert the volumes to liters, because the M unit cancels out. That doesn't feel correct, but it is - the unit cancellation is legal.