Materials: Solution I (1% Compound “A” (m/v); M.W. 800g/mole) Solution II (1.2M
ID: 912744 • Letter: M
Question
Materials:
Solution I (1% Compound “A” (m/v); M.W. 800g/mole)
Solution II (1.2M Compound “B” M.W. 60g/mole; Density of compound “B”: 1.6g/mL)
1. How would you prepare 1mL of each of the following solutions from the above stock solutions.
a. A 1.5mM solution of compound “A”.
b. A 0.36% (m/v) solution of compound “B”.
c. A 6% (v/v) solution of solution I.
d. A solution containing 0.5mg of compound “A” and 0.1% (v/v) of compound “B”.
e. A solution representing the following ratio: solution I: solution II : water : 2:1:2
Explanation / Answer
We have A solution which is 1% m/v. So,
so it means 1 g per 100mL
10 g per liter.
So moles = 10g/800g/mol = 0.0125mol
Molarity = 0.0125mol/ 1L = .0125M
molarity of B = 1.2M
a. M1V1 = M2V2
1.5E-3M (which is 1.5mM) x 1mL = 0.0125M x V2
V2 = 0.12mL
Take 0.12mL of solution I and take water to fill to 1ml.