Answer :
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.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{97.9 g}{58.44 g/mol}\\= 1.675 mol[/tex]
Molarity is the number of moles of a substance present in liter of a solution.
Therefore, volume of given solution is calculated as follows.
[tex]Molarity = \frac{moles}{Volume (in L)}\\3.0 M = \frac{1.675 mol}{Volume}\\Volume = 0.558 L[/tex]
Thus, we can conclude that there are 0.558 liters of the stock solution would she need to have 97.9 grams NaCl.
Final answer:
To prepare a solution with 97.9 grams of NaCl from a 3.0 M stock solution, one would need approximately 0.5587 liters of the stock solution. This is found by converting grams to moles and then dividing by the molarity of the solution.
Explanation:
To calculate the volume of a 3.0 M NaCl stock solution needed to obtain 97.9 grams of NaCl, we first need to find out how many moles of NaCl that corresponds to. We can use the molar mass of NaCl, which is 58.44 g/mol, for this calculation.
First, convert the mass of NaCl needed to moles:
Moles of NaCl = mass of NaCl (in grams) ÷ molar mass of NaCl
Moles of NaCl = 97.9 g ÷ 58.44 g/mol
Moles of NaCl = 1.676 moles
Since we have a stock solution with a concentration of 3.0 M, which means 3.0 moles of NaCl per liter, we can find out the volume needed by dividing the moles of NaCl by the molarity of the solution:
Volume of solution (in liters) = Moles of NaCl ÷ Molarity
Volume of solution = 1.676 moles ÷ 3.0 mol/L
Volume of solution = 0.5587 liters
So, your colleague would need to use approximately 0.5587 liters of the 3.0 M NaCl stock solution to have 97.9 grams of NaCl.
