Hey. I have attempted this multiple times but cannot getall the answers. I guess

Question

Hey. I have attempted this multiple times but cannot getall the answers. I guess I am confused what the 'originalstock solution' is. Can you give me a hand and take methrough this step by step please. Thanks!
You are given 350 mL a stock solution of 1.66E-02 M benzoic acid,MW = 122.125 g/mol.
Answer the following questions.

A. What is the number of moles present in the original stocksolution?
B. You need to dilute the benzoic acid stock solution to 1.24E-03M. If you need 455 mL of solution,
     what volume of theoriginal stock solution do you need to dilute? mL
C. What is the number of moles present in the new dilutedsolution?
D. You mass the solution to be 462.131 g. What is the density ofthe solution? g/mL
E. If the actual density should be 1.05 g/mL. What is your percenterror? % Hey. I have attempted this multiple times but cannot getall the answers. I guess I am confused what the 'originalstock solution' is. Can you give me a hand and take methrough this step by step please. Thanks!

Explanation / Answer

Molarity of the stock solution , M = 1.66 * 10^-2 M Volume of the stock solution , V = 350 mL = 0.35 L A . No . of moles , n = Molarity * Volume in L                                = 5.81 * 10^-3 moles B . MV = M'V' 1.66 * 10^-2 M* V = 1.24 * 10^-3 M * 455 mL                               V = 33.98 mL C. Molarity of the new solution , M' = 1.24 * 10^-3 M     Volume of the new solution , V' = 445 mL =0.455 L No. of moles , n' = M' * V'                          = 0.5642 * 10^ -3 moles D . mass the solution , m = 462.131 g Volume of the new solution , V' = 445 mL Density , d = m /V                 = 1.0384 g / mL E . Percentage error in density = {( difference in thedensities of actual - calculated ) * 100 } / actual density                                               = { ( 1.05 - 1.0384 ) * 100 } / 1.05                                               = 1.1 %    

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