Question

What was the initial concentration of the sucrose solution used in the mini-experiment? Give a step-by-step explanation of how you arrived at this value.

Here are the data points collected:

Empty Dialysis Bag + Clip: 10.55

(V1) Filled Dialysis Bag: 22.98 - 10.55 = 12.43

(V2) Dialysis Bag after 10 minutes: 25.67 - 10.55 = 15.12

Refractive Index: 1.388

The refractive index of the sucrose solution= 1.388

The density equation as given in lab: y=2.6477x-2.5302

density = 1.1448 g/ml

The molarity equation as given is y= 21.019x-28.036

Molarity= 1.138 moles/ liter

Please help me calculate this and explain how you arrived at this result.

Answer 1

The parameters given are density, molarity hence the concentration can be calculated as given below.

concentration of the Using the relation Sucrose Sols can be calculated d=m where, d= density - Moss v=Volume Data given in the problem density = 1.388 molarity = 1-138 msles/tates We see to calculate "Mass" using ndarity by the relation melarity = weight of Solute Molecular wt og solule. Molecular X 1.138 = Weight of solute. weight of (Sucrose) M.W 9 Sucrose os 342.3 g/md. .: 342.3 × 1.138 = Weight of Salute in ltr. 389.53748/tzter weight g Sooves Solute

Since V₂ is the final volume and we know the concentration, the initial concentration or the amount of sucrose in V₁ can be calculated using dilution formulae.

Given that I = 12.43 & v = 15.12. f % Sucrose in 18.10 ml Hence Solution concentration - 8 389.5374 g/1000 ml. 38.953 in 100ml. .. 15.12. = 38.953 x 15.12 Too (final concentra 2 = 5.889 g 1 me - tion q Sucrose ) of sucrole Instial concentration ie 8n 18.43 ml. It can be 3:153 xter3 done using dilution Equation. Nov. = N₂ . N,= x V = 12.43 N = 5-98991-.. Wk. That V = 15.12me.

Hence NV = N. V XX19.43 = 5.889 X 15.12. x = 5:889 XIS.12 12.43 7.1634 g/me Initial concentration = g Sucrose.