Question

You are given a protein solution with a concentration of 0.15 mg/ml.

iii. If the molecular weight of the protein is 22,000 Da, express its initial concentration in moles/liter, micromoles/mL, and micromoles/microliters. If we want 20 micromoles of proteins for a reaction, what volume do we need to prepare?

Answer 1

Answer:

We need  2.933 L of 0.15 mg /mL of protein solution.

Explanation:

Concentration of given solution$C_1 = 0.15 mg/mL$

1 mg = 0.001 g , 1 mL = 0.001 L

$C_1=\frac{0.15\times 0.001 mg}{1\times 0.001 L}=0.15 g/L$

Molecular weight of protein = 22,000 Da =22,000 g/mol

Initial concentration in moles/liter:

$C_1=\frac{0.15 g/L}{22,000 g/mol}=6.8182\times 10^{-6} mol/L$

Initial concentration in micromoles/mL :

1 L = 1000 mL

$C_1=6.8182\times 10^{-6} mol/L=\frac{6.8182\times 10^{-6}\times 10^6 \mu mol}{1000 mL}=6.8182\times 10^{-3} \mu mole/ mL$

Initial concentration in micromoles/microLiter :

1 L = 1000,000 μL

$C_1=6.8182\times 10^{-6} mol/L=\frac{6.8182\times 10^{-6}\times 10^6 \mu mol}{1000000 \mu L}=6.8182\times 10^{-6}\mu mol/\mu L$

Moles of protein required = 20 μmoles

n(Moles)=C(concentration) × V(Volume of solution)

$20 \mu mol=6.8182\times 10^{-6}\mu mol/\mu L\times V$

$V =\frac{20 \mu mol}{6.8182\times 10^{-6}\mu mol/\mu L}$

$V=2.933\times 10^6 \mu L = 2.933 L$

We need  2.933 L of 0.15 mg /mL of protein solution.