Question

A) How many mL of 12.25 M NaOH are required to prepare 525.00 mL of 0.800 M NaOH solution in DI water?

B) How much DI water is required?

Answer 1

A) M1V1 = M2V2

M1 = 0.8 M ; M2 = 12.25 M

V1 = 525 mL ; V2 = ?

0.8 M * 525 mL = 12.25 * V2

V2 = 34.28 mL

M = wt/GMwt *1000/V(mL)

12.25 = x / 40 * 1000 / 34.28

wt of NaOH x = 8.933 gm

Density = mass / volume

Volume of NaOH = mass / density = 8.933 / 2.13 = 4.194 mL NaOH

B) Volume of Dl water = 34.28 - 4.194

                                = 30.086 mL Dl water