LABORATORY 1: MOLARITY AND DILUTION, Lab Reports of Chemistry

Download LABORATORY 1: MOLARITY AND DILUTION and more Lab Reports Chemistry in PDF only on Docsity! LABORATORY 1: MOLARITY AND DILUTION BWJ 10303 SEM II (2022/2023) SECTION 2 DATE: 6/4/2023 INSTRUCTOR: DR. YAP JINGWEI GROUP MEMBERS NO. MATRICULATION 1) NUR AIDA HUSNA BINTI MOHAMAD DESA AW220055 2) PUTERI RIA NORLAILI BINTI ABDUL RAOF AW220053 3) HAIFA HAZWANI BINTI HABIR AW220132 4) HURULAIN SILMINA BINTI HAMDI AW220015 5) MUHAMMAD SUFYAN HARISH BIN SUHAIMI BW180001 INTRODUCTION Molarity is a measure of the amount of solute that is dissolved in a given amount of solution. The number of the moles that solute dissolved in one litre of solution. Therefore, the equation to find the molarity (M) of a solution is M= n/v = moles of solute / litres of solution. The units for this equation are moles per litre. The moles measure the amount of material that you have while molarity measures the concentration of the material. Dilution is a process in which the concentration of a solution is reduced by adding more solvent to the solution which decreases the number of moles per litre (M) where M . V = M . V1 1 2 2 . This experiment typically accurately calculates the molarities for solution. Other than that, it is to prepare a dilute solution from a solution that is more concentrated. Last but not the least, this experiment happens because we want to perform a serial dilution. METHOD Part A: Preparing a 0.10 M sucrose solution in a volumetric flask. 1. The mass of sucrose, C12H22O11 , that required to make 100.0 mL of a 0.10 M solution was calculated. The required amount of sucrose was measured out into a new plastic weighing bottle. 2. The sucrose was transferred into a 100 mL volumetric flask and using a wash bottle to rinse any of the solid remaining on the weighing boat. The water was added to the flask until it was one-half to two thirds full. 10 drops of food colouring to the solution were added. 3. The flask was then capped and inverted gently several times until the solid dissolves completely. A thumb or forefinger was used to secure the cap onto the flask while inverting. Do not shake the flask hard as the glass neck may break. 4. Distilled water was added slowly until the water level was near the etched line on the neck of the flask. More water was added carefully, drop by drop until the bottom of the meniscus was on the etched line. 5. The flask was then capped and inverted gently 3 to 4 more times. Do not shake the flask hard as the glass neck may break. 6. The solution was transferred to the labelled plastic bottle once it was thoroughly mixed and all solid has been dissolved. The bottle was capped to prevent contamination or evaporation. This was the 0.10 M sucrose stock solution. 7. Step 1-6 were repeated for the remaining concentration ( 0.20 M, 0.30 M, 0.4 M, 0.5 M, 0.6 M) Figure 1: 0.10M of sucrose solution Table 2 Test tube 1 Test tube 2 Test tube 3 Test tube 4 Test tube 5 M1 × V1 = M2 × V2 Concentration of concentrated solution (M) 0.20 0.20 0.02 0.002 0.0002 Volume(in mL) of concentrated solution added 10.0 1.00 1.00 1.00 1.00 Concentration of Dilute solution (M) 0.20 0.02 0.002 0.0002 0.00002 Volume(in mL) of Dilute solution 10.00 10.00 10.00 10.00 10.00 Colour observations Rank solution 1-5 lightest = 1 darkest=5 5 4 3 2 1 Figure 2: 0.20M of sucrose solution Table 4 Test tube 1 Test tube 2 Test tube 3 Test tube 4 Test tube 5 M1 × V1 = M2 × V2 Concentration of concentrated solution (M) 0.40 0.40 0.04 0.004 0.0004 Volume(in mL) of concentrated solution added 10.00 1.00 1.00 1.00 1.00 Concentration of Dilute solution (M) 0.40 0.04 0.004 0.0004 0.00004 Volume(in mL) of Dilute solution 10.00 10.00 10.00 10.00 10.00 Colour observations Rank solution 1-5 lightest = 1 darkest=5 5 4 3 2 1 Figure 4: 0.40M of sucrose solution Table 5 Test tube 1 Test tube 2 Test tube 3 Test tube 4 Test tube 5 M1 × V1 = M2 × V2 Concentration of concentrated solution(M) 0.50 0.50 0.05 0.005 0.0005 Volume(in mL) of concentrated solution added 10.00 1.00 1.00 1.00 1.00 Concentration of Dilute solution (M) 0.50 0.05 0.005 0.0005 0.00005 Volume(in mL) of Dilute solution 10.00 10.00 10.00 10.00 10.00 Colour observations Rank solution 1-5 lightest = 1 darkest=5 5 4 3 2 1 Figure 6: 0.60M of sucrose solution CALCULATION A. The mass of sucrose, C12H22O11 that required for each molarity of 0.10M, 0.20M, 0.30M, 0.40M, 0.50M and 0.60M sucrose solution: 0.10M solution: C12H22O11= (12 12.01) + (22 1) + (11 16) = 342.12g× × × Mass = 𝑥 Volume = = 0.1L100𝑚𝐿 1000𝑚𝐿 Molarity = 0.10M Molarity = Moles =𝑀𝑜𝑙𝑒 𝑉𝑜𝑙𝑢𝑚𝑒 (𝐿) 𝑀𝑎𝑠𝑠 (𝑔) 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑔/𝑚𝑜𝑙) 0.10M = 0.01 mol =𝑀𝑜𝑙𝑒 0.1𝐿 𝑀𝑎𝑠𝑠 (𝑔) 342.12 (𝑔/𝑚𝑜𝑙) Mole = 0.01 mol Mass = 3.4212g 0.2M solution : C12H22O11= (12 12.01) + (22 1) + (11 16) = 342.12g× × × Mass = x Volume(V)= = 0.1L100𝑚𝐿 1000𝑚𝐿 Molarity (M)= 0.20M Molarity = Moles =𝑀𝑜𝑙𝑒 𝑉𝑜𝑙𝑢𝑚𝑒 (𝐿) 𝑀𝑎𝑠𝑠 (𝑔) 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑔/𝑚𝑜𝑙) 0.20M = 0.02 mol =𝑀𝑜𝑙𝑒 0.1𝐿 𝑀𝑎𝑠𝑠 (𝑔) 342.12 (𝑔/𝑚𝑜𝑙) Mole = 0.02 mol Mass = 6.8424g 0.3M solution: C12H22O11= (12 12.01) + (22 1) + (11 16) = 342.12g× × × Mass = x Volume(V)= = 0.1L100𝑚𝐿 1000𝑚𝐿 Molarity (M)= 0.30M Molarity = Moles =𝑀𝑜𝑙𝑒 𝑉𝑜𝑙𝑢𝑚𝑒 (𝐿) 𝑀𝑎𝑠𝑠 (𝑔) 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑔/𝑚𝑜𝑙) 0.30M = 0.03 mol =𝑀𝑜𝑙𝑒 0.1𝐿 𝑀𝑎𝑠𝑠 (𝑔) 342.12 (𝑔/𝑚𝑜𝑙) Mole = 0.03 mol Mass = 10.2636g 0.4M solution C12H22O11= (12 12.01) + (22 1) + (11 16) = 342.12g× × × Mass = x Volume(V)= = 0.1L100𝑚𝐿 1000𝑚𝐿 Molarity (M)= 0.40M Molarity = Moles =𝑀𝑜𝑙𝑒 𝑉𝑜𝑙𝑢𝑚𝑒 (𝐿) 𝑀𝑎𝑠𝑠 (𝑔) 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑔/𝑚𝑜𝑙) 0.40M = 0.04 mol =𝑀𝑜𝑙𝑒 0.1𝐿 𝑀𝑎𝑠𝑠 (𝑔) 342.12 (𝑔/𝑚𝑜𝑙) Mole = 0.04 mol Mass = 13.6848g 0.20 M Vesa Wee 1 Mm, = 0-00 M Myvi = Ma Vo Vv, « 10:00 0-2 (10) = M2 Cto) +2 Ms 2 2 Ms ? 7: V> = 10:00 20-20M Tesi wwe 9 M, = o-20™ MiNi > Ma Va yy 2 roo 0-00) = macro) Ms = 2 Ms 20:08™ Va > 10-60 Jest wre 3 M)= 0:02 ™ Mv = Mo Vo Vy = 00 oro (t) = Ma Cto) Moe Ms = 0.00TM Vo 210-00 0.30 M e Test tube 1 Mi = 030M M2 7M Vy 21000 mL — yg = \0.00mb MV, = MaVo (aon ZoM) (\0.6OmL) = Mz (10.00 My = (230M) C160 Jest ee My = 0-009 Mw = MaVe wy = y1200 9-0 080) = Malte) Maz? Ms 2 0-0003M No 2 19-00 Jesi tive © = Mave My > 6,0008 maven v, 21.0008 Oe Sus 5 Ma 2e00008™ Vp = 10 -GOCo ● Test tube 2 ● Test tube 3 ● Test tube 4 ● Test tube 5 M, = 0,0003M Vi = 1 OOmL MV =Ma Vo CO.0003MYCNOOmLY = M2 (\0.00m) 0.40 M 2x Test avioe | Mis o-gom Viz 10rOombl Ma: 9 Vz 2 \O:00 me Miv = Movy (O40) (10°00) = Ma (10-00) s>Co ho) (o-oo) (10-00) Mo: o-4o mM & Tesh sve Mi, = 0-004 M soo mL Vie M32 7 Va = \Or00 mL My Vy 2 Mov (o-00¥) (10°) = mz (\000) ms = (000) (00) cio eo) Mat 0-000 YM Tess ave 2 mM, o fom Vit (00 me Mae 7 Vo=10:00m\ MW, eM oN5 Coro) (400) = Ma (10-00) M5: C1000) Maz gro4¥m B® Test tue 5 Mz 0-000} M Vit yO mL M21 No =O 00mL Myvi = Movo a Tess tue 3 ™,= 0-04M Viz oO mL Ma: 7 Na = 10°00 mb Mi = M3 V5 (0°04) (\100) = M3 (10-00) My > O$ D100) (000) M27 O'00k MH €0:000y4 ) C1100) = ma (10-00) Ma 2 (0: 000% )(\100) C\o08) M3 = 0°0000 4 Mm DISCUSSION Sucrose is a naturally occurring sugar found in various amounts in plants like fruits, vegetables and nuts. Sucrose dissolves in water because of the characteristic of water that is polar molecules which attract the negative and positive areas on the polar sucrose molecules which makes the sucrose manage to dissolve in water. Molarity also known as concentration is the number of moles of a substance per litre of solution. Solutions labelled with the molar concentration are denoted with a capital M. A 1.0 M solution contains 1 mole of solute per litre of solution. Concentration, volume and moles are connected in this equation: 𝑀 = 𝑛 𝑉 Where: M = Molarity (mol/L) n = mol V = volume And moles can be found by this equation: 𝑚𝑜𝑙 = 𝑚𝑎𝑠𝑠 𝑀𝑟 The first part of the lab experiment was to prepare a sucrose solution with different molarity. Starting from 0.10 M to 0.60 M. Each different molarity gives a different mass of sucrose solvent that needs to be calculated using the molarity formula then with the mol formula as above. Different molarity gives a different concentration of the sucrose solution. Based on the experiment that had been done, the increase of the molarity, the more mass of sucrose solvent needed to make the sucrose solution. Dilution refers to the process of adding additional solvent to a solution to decrease its concentration. This process keeps the amount of solute constant, but increases the total amount of solutions, thereby decreasing its final concentration with an identical solution of lesser concentration. Diluting solutions is a necessary process as stock solutions are often purchased and stored in very concentrated forms. The volume of solvent needed to prepare the desired concentration of a new, diluted solution can be calculated mathematically. The formula is as followed: M1V1= M2V2 M1 denotes the concentration of the original solution and V1 denotes the volume of the original solution; M2 represents the concentration of the diluted solution, and V2 represents the final volume of the diluted solution. When calculating dilution factors, it is important that the units for both volume and concentration are the same for both sides of the equation. During the dilution process which in part B, all the sucrose solution from different molarity needs to undergo dilution process mix with distilled water to observe the concentration from the first test tube until the fifth test tube. Based on the experiment that was done from the 0.10 M to 0.60 M sucrose solution, the concentration of the sucrose solution is getting lesser and lesser which causes the colour to change slightly from darker to less darker as expected before the experiment begins. This happened because the relative intensity of colour is proportional to the concentration of the dissolved compound. The greater the compound's concentration, the darker the solution colour appears and vice versa. In the next laboratory experiment for the future, it will be better if 2mL of sucrose solution is being taken from every test tube of sucrose solution to observe the drastic changes of the colour from test tube #1 to test tube #5. CONCLUSION In this lab, we can conclude that the molarity of a solution can be observed through its sight as molarity and colour are related. As the molarity of a solution increases so does the intensity of a colour, and as the molarity of a solution decreases the colour intensity also decreases. We can also conclude that the solution can be diluted by adding more solvent to lower the molarity. From this lab, it can also be concluded that when using a pipet not to plough the solution back into its container otherwise, it will go everywhere. QUESTIONS 1. What is the molar mass of sucrose, C12H22O11? 2. How many grams of sucrose would be needed to prepare 100 mL of a solution of 0.10 M sucrose? Show your calculations. 3. Calculate the volume of 0.10 M sucrose solution, in mL, that must be diluted to prepare 10.00 mL of a 0.050 M sucrose solution. Show your work.