Question

4. The concentration of both molecules in the starting mixture, prior to separation, is 1 mM. What is the amount (in mass units such as g) of each molecule were present in 200 L of the sample mixture? You must show all of your work to obtain credit.

Answer 1

Answer:

Hi there, the question here is not complete. Not to worry there will be in step by step guild in the explanation section that will help you in solving similar question.

Explanation:

So, an important information that should have been given in this question is the same of each of molecules in the mixture or even the name or identity of the mixture.

Nonetheless, the molarity if the mixture is given to be 1mM and the volume of the mixture is given to be 200L.

Therefore, the first thing to do is to determine the number of moles of each molecules in the mixture. After that the mad of each molecules can be determined using the number of moles gotten.

Thus, 1 mM = 0.001 M. Therefore, the number of moles of the molecules in the mixture can be calculated by using the formula below:

Number of moles = molarity × Volume = 0.001 × 200 = 0.2 mol.

Hence the mass of each Molecules can be calculated by using the formula below;

Number of moles = Mass/ molar mass.

That's mass = number of moles × molar mass of each molecules.

Assuming the molecules in the mixture are A and B and the respective molar mass of A and B are x g/mol and y g/mol. We will have;

The mass of A = 0.2 moles × x g/mol.

The mass of B = 0.2 moles × y g/mol.