Question

Write how to prepare the following:

1. 20.0 mL of 25 mM lysine from solid

2. 1 L of 0.5 M of NaCl from a 6 M stock of HCl

3. 10 mL of 0.60 M HCl from a 6 M Stock HCl

Answer 1

(1) Molar mass of lysine is 146.2 g/mol

We know that Molarity , M = ( mass/molar mass) x ( 1000/Volume in mL)

25 mMx10^-3 (M/mM) = ( mass/146.2) x ( 1000 / 20.0 mL)

mass = 0.073 g

So exactly weigh and transfer 0.073 g of lysine and add 20.0 mL of distilled water and allowed to dissolve all the lysine we get required solution

(2) According to law of dilution   MV = M'V'

Where M = Molarity of stock = 6 M

V = Volume of the stock = ?

M' = Molarity of dilute solution = 0.5 M

V' = Volume of the dilute solution = 1L = 1000 mL

Plug the values we get , V = ( M'V') / M

   = 83.3 mL

So 83.3 mL of the stock solution was taken and make up the solution to 1L by distilled water we get required solution.

(3) According to law of dilution   MV = M'V'

Where M = Molarity of stock = 6 M

V = Volume of the stock = ?

M' = Molarity of dilute solution = 0.60 M

V' = Volume of the dilute solution = 10 mL

Plug the values we get , V = ( M'V') / M

   = 1.0 mL

So 1.0 mL of the stock solution was taken and make up the solution to 10 mL by using distilled water we get required solution.