If I have 340 mL of a 0.5 M NaBr solution, what will the concentration be if I add
400 mL more water to it?
A. 0.03 molar
B. 0.13 molar
C. 0.23 molar
D. 0.33 molar

Option C. 0.23 molar

Explanation:

The following data were obtained from the question given:

C1 = 0.5 M

V1 = 340mL

V2 = since 400mL was added to the original volume, therefore = 340 + 400 = 740mL

C2 =?

The concentration of the diluted solution can be calculated by using the following:

C1V1 = C2V2

0.5 x 340 = C2 x 740

Divide both side by 740

C2 = (0.5 x 340) / 740

C2 = 0.23M

Therefore, the new concentration of the solution is 0.23M

1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles