A stock solution has a concentration of 1.5 M NaCl and is diluted to a 0.80 M solution with a volume of 0.10 L. What volume of the stock solution in liters was used to make the diluted solution

0.053 L  is the volume of concentrated solution that was used

Explanation:

Let's determine the answer of this, by rules of three.

There is also a dilution formula.

Molarity is a sort of concentration that indicates the moles of solute in 1L of solution.

In 1 L of concentrated solution, there are 1.5 moles of NaCl

In 1 L of diluted solution, there are 0.80 moles.

The volume for the diluted solution is 0.10L

The rule of three will be:

1L of solution has 0.80 moles of solute

Then, 0.10L of solution must have (0.1 . 0.8)/1 = 0.08 moles

This moles came from the concentrated solution, and we know that in 1L of this solution we have 1.5 moles. Therefore the rule of three will be:

1.5 moles are in 1L of solution

0.08 moles were in (0.08 . 1L / 1.5) = 0.053 L (This is the volume of concentrated solution that was used)

Dilution formula is: M conc . Vol conc = M diluted . Vol diluted

1.5 M . Vol conc = 0.80 M . 0.10L

Vol conc = 0.80 M . 0.10L / 1.5M = 0.053L

0.053 L of stock solution was used.

Explanation:

Step 1: Data given

Concentration of the stock solution = 1.5 M NaCl

Concentration of the diluted solution = 0.80 M

Volume of the diluted solution = 0.10 L

Step 2: Calculate the volume of the stock solution

C1*V1 = C2*V2

⇒with C1 = the concentration of the stock solution = 1.5 M NaCl

⇒with V1 = the volume of the stock solution = TO DETERMINED

⇒with C2 = the concentration of the diluted solution = 0.80 M

⇒with V2 = the volume of the diluted solution = 0.10 L

1.5 M * V1 = 0.80 M * 0.10 L

V1 = (0.80 *0.10 L)/ 1.5 M

V1 = 0.053 L

0.053 L of stock solution was used.

the answer would be b i believe