I have worked part a but I have no clue how to do the rest w

5.) Dilutions a. You find a solution of 255 mg/mL cuso4. Is this the same as a 20% (w/v)solution? If not, how would you alter the solution to arrive at the desired concentration? b. You have a solution of 2M c6H1206. Is this the same as a 10% (w/v) solution? If not, how would you alter the solution to arrive at the desired concentration? c: You have a solution of 25 mM NaN3. ls this the same as a 15 ng/mL solution? If not, how would you alter the solution to arrive at the desired concentration? If not, how d. You have a solution of 4.59tchCIN. s this the same as a 150 mM solution would you alter the solution to arrive at the desired concentration? e. You find a solution of 103 ng/uL c21H20BrN3. Is this the same as an 8o mM solution? If not, how would you alter the solution to arrive at the desired concentration?

Solution

b) Molecular weight of C6H12O6 = 180.1559 g/mol

% w/v = (mass solute in g / volume of solution in ml) x 100

10 % w/v means 10g/100 ml

Molarity is expressed as M or mol/L

We have 2M solution which means 2mol/L

2 mol/L x 180.1559g/mol x 1L/10³ ml = 0.36 g/ml

To express in %w/v, multiply numerator and denominator by 100

36.03 g/ 100ml = 36.03% w/v

So 2M is not equal to 10% w/v

To get the desired concentration,

v1 x c1 = v2 x c2

v1 x 10 = 100 x 36.03

v1 = 360.3 ml

Add 360.3 ml of solvent to 2M solution to get 10% w/v

c) Molecular weight of NaN3 = 65.0099 g/mol

25 x 10^-3 mol/L x 65.0099g/mol x L/10³ ml = 1.625 x 10^-3 g/ml = 1.625 mg/ml

So 25mM is not equal to 15ng/ml

To get the desired concentration

C1 x v1 = c2 x v2

15 x 10^-9 g x v1 = 1.625 x 10^-3 g x ml

v1 = 108.33 x 10³ ml = 108.33 liter

Add 108.33 liter solvent to the 25mM solution to get the desired concentration of 15ng/ml