Question

1. To prepare 20 mL of 15 mM NaCl, how much NaCl would you need to weigh?

2. To prepare 5 mL of 1 mM glucose, how much glucose would you need to weigh? How do you prepare 1 L of 1 mM NaCl using the NaCl solution you prepared in (1)?

3. How do you prepare 1 mL of 10 µM glucose using the glucose solution you prepared in (2)?

Answer 1

1) Molarity = (W/MW) (1000 / v in mL)

0.015 = (W/58.5) (1000 / 20)

W / 58.5 = 0.0003

W = 0.0175 g

mass of NaCl required = 0.0175 g

2) 0.001 = (W/180) (1000 / 5)

W/180 = 0.000028

W = 0.00504 g

mass of glucose required = 0.00504 g

part 2 in 2)

M1V1 = M2V2

M1 = 0.015 , V1 = ?

M2 = 1mM = 0.001 , V2 = 1000 mL

V1 = (M2V2 / M1) = (0.001 x 1000 / 0.015)

V1 = 66.7

66.7 mL solution prepared in 1 should be dilute to 1 L with water

3) M1 = 0.001 M , V1 = ?

M2 = 0.01 M , V2 = 1.0 mL

V1 = (M2V2 / M1) = (0.01 x 1.0 / 0.001)

V1 = 10 mL

10 mL solution in part 2 must be evapourate solvent up to 1 mL