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Let’s say a colleague working in the lab needs to create a solution containing 97.9 grams of NaCl. If she has a 3.0 M stock solution of NaCl dissolved in water, how many liters of the stock solution would she need to have 97.9 grams NaCl? Remember the molar mass of NaCl is 58.44 g/mol.

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Answer: There are 0.558 liters of the stock solution would she need to have 97.9 grams NaCl.

Step-by-step explanation:

Given: Molarity = 3.0 M

Mass of NaCl = 97.9 g

Molar mass of NaCl = 58.44 g/mol

As number of moles is the mass of substance divided by its molar mass.

So, moles of NaCl are calculated as follows.


Moles = (mass)/(molar mass)\\= (97.9 g)/(58.44 g/mol)\\= 1.675 mol

Molarity is the number of moles of a substance present in liter of a solution.

Therefore, volume of given solution is calculated as follows.


Molarity = (moles)/(Volume (in L))\\3.0 M = (1.675 mol)/(Volume)\\Volume = 0.558 L

Thus, we can conclude that there are 0.558 liters of the stock solution would she need to have 97.9 grams NaCl.

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