Question

A 931 ml nacl solution is diluted to a volume of 1.34 l and a concentration of 2.00 m . what was the initial concentration?

A) 0.621 M
B) 1.29 M
C) 1.51 M
D) 3.43 M

Answer 1

To find the initial concentration of a diluted NaCl solution, we use the dilution formula: M1V1 = M2V2. By substituting the given values into the formula, we find that the initial concentration was approximately 3.09 M, correct option is not given.

Step-by-step explanation:

The student is asking how to calculate the initial concentration of a sodium chloride (NaCl) solution before it was diluted. The initial volume of the NaCl solution was 931 mL (which is 0.931 L, since 1000 mL = 1 L), and after dilution, the final volume was 1.34 L at a concentration of 2.00 M (molar). To find the initial concentration, we can use the dilution formula:

M1V1 = M2V2

Where:

• M1 is the initial molarity
• V1 is the initial volume
• M2 is the final molarity (2.00 M)
• V2 is the final volume (1.34 L)

Rearranging the formula to solve for M1 gives us:

M1 = (M2V2) / V1

Substitute the known values:

M1 = (2.00 M * 1.34 L) / 0.931 L

Calculating this we get:

M1 = 2.876 / 0.931

M1 = 3.09 M

Therefore, the initial concentration of the NaCl solution was approximately 3.09 M. However, this value is not listed among the available choices, and there may have been a typo in the student's question or the answer choices provided.