.401g of ASA (180.16g/mol) is diluted with water into a 250-mL volumetric flask

Question

.401g of ASA (180.16g/mol) is diluted with water into a 250-mL volumetric flask (Solution A).  50-mL of this solution is then transfered into a new 250-mL volumetric flask and diluted with water (Solution B).  Next 5-mL of solution B is transfered into a 100-mL volumetric flask, what is the concentration of the final solution? is this correct ?

my work:

.401g/180.16g/mol = 2.23x10^-3 mol ASA

Concentration of Solution A = 2.23x10^-3 mol ASA/.25L = 8.9x10^-3 M

Concentration of Solution B = (8.9x10^-3 M)(.05L)/.25L = 1.78x10^-3 M

Final Concentration = (1.78x10^-3 M)(.005L)/.1 = 8.9x10^-5

Explanation / Answer

Yes my friend... its correct :)

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