Question

describe how to make one liter of the following solution: 0.3 M mannitol (FW=182), 0.02 M KPO4, 0.01 M KCl, 0.005 M MgCl2. You have the following stocks... 1 M KPO4, 1 M KCl, 1 M MgCl2.

Answer 1

1. Pepare 0.3 M mannitol

Molarity = weight * 1000 / molecular weight * V in ml

Substitute values in above equation, we get

0.3 = weight * 1000 / 182 * 1000 ml (1l = 1000 ml)

weight = 182* 0.3 = 54.6g

Dissolve 54.6 g of mannitol in 1L of water to get 0.3M mannitol.

2.Prepare 0.02 M KPO4 from stock 1M KPO4

We know C1V1 = C2V2 ( Where C1,C2 are initial and final concentrations respectively and V1 ,V2 are initial and final volumes of the solutions respectively),

So, we get 1M* V1 = 0.02 * 1L

V1 = 0.02L

=20 ml

Take 20 ml of KPO4 from stock 1L and (add 1000-20 )980 ml of water to make up the volume to 1L.This is 0.02M KPO4.

3.Prepare 0.01 M KCl from stock 1M KCl

C1V1 = C2V2

1M*V1 = 0.01 * 1L

V1 = 0.01L

=10 ml

Take 10 ml of KCl from stock 1L and (add 1000-10 )990 ml of water to make up the volume to 1L.This is 0.01M KCl

4.Prepare 0.005 M MgCl2 from stock 1M MgCl2

C1V1 = C2V2

1M*V1 = 0.005 * 1L

V1 = 0.005L

=5ml

Take 5 ml of MgCl2 from stock 1L and (add 1000-5 )995 ml of water to make up the volume to 1L.This is 0.005 M MgCl2