Serial dilutions. Need help with my lab

Question

PART A 

 

Mass of empty weighing dish=2.0795g
Mass of weighing dish+NaOH=2.1808g
Mass of NaOH=0.1013g
Final Volume of NaOH stock solution A=50mL
Concentration of NaOH stock solution = ?

For that part^ , how do I find the concentration of NaOH stock solution A?

PART B

Beaker label B
Volume of preceding solution taken=35mL
Initial Concentration of NaOH solution(M)=?
Moles of NaOH in source solution and final solution=?
Volume of water dispensed from the burette : Final burette reading =15mL ,Initial burette reading=0mL , volume used=15mL
Final concentration of solution(M) = ?

For that part ^ , how do I find the moles of NaOH in the source and final solution, Initial concentration of NaOH solution, and the final concentration of solution(M)?

Answer 1

Part A - Determine the number of moles of NaOH from its mass (0.1013 g). The concentration in molar units is given by:

 

M = (moles solute)/(Liters of solution)

 

Part B - The number of moles in the solution is simply given by multiplying concentration with volume:

 

(moles of sample) = (molarity)x(volume in liters)

 

I'm not sure what you mean by dispensed burette volume. Is that the amount of NaOH dispensed or amount of water added? The number of moles of NaOH will not be the same for each case.