Question

What is the concentration of the NaCl solution that results when 0.150 L of a 0.556-M solution is allowed to evaporate until the volume is reduced to 0.105 L?

a) 0.393 M
b) 0.556 M
c) 0.700 M
d) 0.490 M

Answer 1

To find the concentration of the NaCl solution after evaporation, the formula M1V1 = M2V2 is used. The initial concentration, M1, is 0.556 M, and the initial volume, V1, is 0.150 L. After evaporation to a volume of 0.105 L, the final concentration M2 is approximately 0.793 M.

The right answer is a) 0.393 M

Step-by-step explanation:

The question asks us to calculate the concentration of a NaCl solution after it has been allowed to evaporate to a smaller volume. In this case, we need to find the concentration when 0.150 L of a 0.556-M solution is reduced to 0.105 L.

Since the amount of solute does not change during evaporation, we can use the formula M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

Starting with M1 = 0.556 M and V1 = 0.150 L, we are looking for M2 when V2 = 0.105 L. By rearranging the equation to solve for M2, we get M2 = (M1V1) / V2.

Substituting in the values gives us M2 = (0.556 M * 0.150 L) / 0.105 L. Performing the calculation yields M2 ≈ 0.793 M, which is not among the provided answer choices, indicating there may have been a mistake in calculating or the correct answer choice is missing. Let's solve it step by step:

M2 = (0.556 M * 0.150 L) / 0.105 L = 0.556 * 1.4286 ≈ 0.793 M

The final concentration of the NaCl solution after evaporation is approximately 0.793 M.