Question

375 mL of a 0.455 M sodium chloride (NaCl) solution is diluted with 1.88 L of water. What is the new concentration in molarity?

Answer 1

Answer:

= 0.0908 M

Explanation:

Using the dilution formula;

M1V1 = M2V2

where M1 is the initial concentration, V1 is the initial volume, M2 is the concentration attained while V2 is the new volume.

Therefore;

M2 = M1V1/V2

M1 = 0.455 M

V1 = 375 mL or 0.375 L

V2 = 1.88 L

Thus;

M2 = (0.455 × 0.375)/1.88

    = 0.0908 M

Answer 2

Answer:

0.0757 M

Explanation:

The question asks for the new concentration of a solution after being diluted. For this we can use the dilution formula:

M₁ V₁ = M₂ V₂

Where M stands for molarity and V for volume, so M₁ and V₁ are the initial values for molarity and volume, and  M₂ and V₂ are the final values of molarity and concentration.

So

M₁ = 0.455 M

V₁ = 375 mL = 0.375 L (because 1 L  = 1000 mL)

The question says the solution is diluted with 1.88 L of water, so to the initial 375 mL of solution, 1.88L of water are added, hence the final volume is:

V₂ = 0.375 L + 1.88L = 2.255 L

M₂ is the final concentration which we want to calculate, so from the formula we get:

$M_{2} = \frac{M_{1} V_ {1}}{V_{2}}$

M₂ = (0.455 M × 0.375 L) / 2.255 L = 0.0757 M

The new concentration after the dilution is 0.0757 M