Question

i need help creatimg a graph and solving to fill in all the blanks. please help me out and show me how to solve to fill in the blanks

Part 1: Determining molar conductivity of NaCI Stock Solution Stock NaCI solution Supporting Calculation |none required none required Mass of NaCI Volume of solution Molarity of stock solution 8,80% 256 MC Diluted Standards Standard mL of stockM Conductivity mS/cm Molarity Supporting calculation for molarity diluted to 100 mL moles/L 5.00 280, 3 286.0 2 10.00 15.00 20.00 244.8 3013 25.00 303 0 151

Answer 1

Molarity of stock solution

M1 = (8.80g/58.45 g/mol)*(1/0.250 L)

= 0.602 M

For dilution process M1V1= M2V2

M2 = M1V1/V2

M1= 0.602 M and V2= 100 mL

Standard 1 V1 = 5.00 mL

Molarity of standard 1, M2 = M1V1/V2

= (0.602M*5.00mL)/100 mL

= 0.301 M

Malarity of standard 2

M2= (0.602 M*10.00mL)/100 mL

= 0.0602 M

Molarity of standard 3

M2 = (0.602 M*15.0mL)/100 mL

= 0.0903 M

Molarity of standard 4

M2= (0.602 M*20.0 mL)/100 mL

= 0.120 M

Molarity of standard 5

M2= (0.602 M*25.0 mL)/100 mL

= 0.151 M