Question

PLEASE HELP
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.

Answer 1

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

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.

$Moles = \frac{mass}{molar mass}\\= \frac{97.9 g}{58.44 g/mol}\\= 1.675 mol$

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

Therefore, volume of given solution is calculated as follows.

$Molarity = \frac{moles}{Volume (in L)}\\3.0 M = \frac{1.675 mol}{Volume}\\Volume = 0.558 L$

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