Question

If 300ml of 1.2 M NaCl is diluted to 450 ml, what is the final molarity of NaCl?

a) 0.8 M
b) 1.0 M
c) 1.5 M

Answer 1

The final molarity of a solution originally at 1.2 M NaCl diluted from 300 ml to 450 ml is 0.8 M, calculated using the dilution equation M1V1 = M2V2.

Step-by-step explanation:

If 300ml of 1.2 M NaCl is diluted to 450 ml, the final molarity of NaCl can be calculated using the dilution 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 of the solution.

Using the given values:
M1 = 1.2 M (initial molarity)
V1 = 300 ml (initial volume)
V2 = 450 ml (final volume)

We need to find M2 (final molarity).

M1V1 = M2V2
1.2 M x 300 ml = M2 x 450 ml
M2 = (1.2 M x 300 ml) / 450 ml
M2 = 0.8 M

Therefore, the final molarity of the NaCl solution after dilution is 0.8 M.