You see the concentration decreased ten fold, from 1 M to 0.1 M.You can simply increase the volume by ten times. For exampleIf you have 250 mL of 1 molar HCl, you can add distilled water upto 2500 mL. Now the concentration is 0.1 molar. If you want to use the serial dilution method, you can use the C1V1=C2V2 equation.C1 = starting concentration.C2 = final concentration.V is for volume.So let's say you have 500 mL of 1 M HCl and you only want 50 mL of 0.1 M HCl.1 M x V1 = 0.1 M x 50 mL V1 = 5 mL So you take 5 mL of the original (stock) solution and dilute it up to 50 mL. Then you have 50 mL of 0.1 M HCl.
Describe how you would prepare 2.00 L of 0.20 M NaOH solution from solid NaOH.?
You have a solution of table salt, NaC1, in water. What happens to the salt concentration ( increases, decreases, or stays the same) as the solution boils? Breifly explain how you arrived with your answer 5) Calculate the concentrations of all the ions present in each of the following solutions of strong electrolytes. a. 0.25 M CaC12 b. 0.15 M A1(NO3)3 c. 42.4 g K3PO4 in 250.0 mL of solution Describe how you would prepare 2.00 L of 0.20 M NaOH solution from solid NaOH. 1) a solution of KNO3 is prepared by dissolving 6.0 g of KNO3 crystals in 14.0 g of deionized water. a. What is the solute in the KNO3 solution? b. What is the solvent in the KNO3 solution? c. What is the mass percent of KNO3 in the solution?
We know that molarity is the no of moles per unit volume.From following relation we can calculate amount of cacl2 required (X)to prepare 0.25 M cacl2 250 ml solution.1000ml—1M—111g cacl2250ml —0.25—X g cacl2So that from above1000ml ×1M×X g= 250ml×0.25M×111gX=[250ml×0.25M×111g]÷[1000ml×1M]X=6.9375 g of cacl2Hence by dissolving 6.9375 g of cacl2 in 250 ml volumetric flask with distilled water then you will get 0.25 M cacl2 solution.
A2A.Easiest way is to use C1V1 = C2V2C1 = stock concentrationV1 = stock volumeC2 = final concentration V2 = final volumeYou can use any units, doesn't have to be SI units, as long as you keep it consistence (i.e. use the same units on both sides of the equation.)So,3.0 M x V1 = 1.0 M x 100 mLV1 = 33.33 mLThus you can take 33.33 mL of the stock solution and dilute it up to 100 mL to get a 1.0 M solution.You may adjust the final volume a little bit so that the amount you need to measure from the stock solution is a rounded figure. For instance, if you're preparing 150 mL instead of a 100 mL, you will measure out 50 mL from the stock solution instead. This would give you a more accurate solution. Although it's not difficult to measure out small volumes with a micro pipette either.You can also easily recognize that this is a three fold dilution. I just gave you the formula for those that are not very straight forward.
How would you prepare 100ml of 0.1M NaCl solution?from the prepared solution how to prepare 10ml of 10mM NaCl?
For the best answers, search on this site https://shorturl.im/awwds Easily, Molarity is Moles of Solute (NaCl) divided by the number of litres of solution (H2O) therefore, You begin with the equation 0.1M = x(Moles of Solute)/0.1(Litres) and solve for "x" 0.01mol =x Then, you convert 0.01mol of NaCl into grams weight in grams/0.01mol=58.5grams per mole/1mol weight in grams=.588g so you should mix .588 grams of NaCl with 100 mL of water.
The standard solution of FeSCN2+ (prepared by combining 9.00 mL of 0.200 M Fe(NO3)3 w/1.00 mL of 0.0020 M KSCN
To the previous answerer: good job copying and pasting, YOU DIDNT ANSWER THE QUESTION. look to the previous answerer's post for the theory. here's how to answer it: equations needed: [SCN–]eq = [SCN–]i – [FeNCS2+]eq [FeNCS2+]eq = (Aeq/Astd) * [FeNCS2+]std [SCN–]i is given= .00070 M Aeq is given= .150 Astd is given= .530 [FeNCS2+]std = .0002 plug it in! [FeNCS2+]eq = (.150/.530) * .0002 = .000056604 M [SCN–]eq = .00070 M - .000056604 M = .0006434 M <-- answer
Given volume of acid (HCl) is 50 ml and concentration is 0.4 NVolume of base (NaoH) is 50ml and concentration is 0.2 N .Actually it is simple reaction in which salt is formed then some acid is remains find pH of that acid .Base is 0.2 * 50Nml = 10NmlAcid is 0.4*50Nml = 20Nml.So remaining acid is 20–10 = 10Nml in 100ml of solution. So concentration of acid is 10/100N = 0.1NSo pH is -log[H+]pH = -log(0.1)= 1.Therefore pH of solution is 1
How do you prepare 500ml of a 3M NaCl solution?
3M means 3 moles of NaCl per liter. Recall that 1000 ml - 1 L, so you have 1.5 moles NaCl per 500 ml 1 mole of NaCl is roughly 58 g, so 1.5 moles is 87g of NaCl Dissolve 87g of NaCl in 500 ml of water and there u have it. (you should be able to do this kind of simple problem in your head, imo, especially if it's a multiple choice exam...you can just estimate your answer, I knew that Na molar mass is 23 and Cl is 35)
NaCl dissociates in a 1:1:1 ratio according to the equation:NaCl --> Na+ + Cl-So, based on this ratio if you have 0.2 molar NaCl you will get 0.2 molar Na+ and Cl-. The 1:1:1 ratio means that the molarity of each of the ions will be the same as the molarity of NaCl.Contrast this to the situation for MgCl2. This dissociates in a 1:1:2 ratio:MgCl2 --> Mg2+ + 2Cl-In this case if you have 0.2 molar MgCl2, it will dissociate to give 0.2 molar Mg2+ but 0.4 molar Cl- because there are twice as many Cl- ions as MgCl2 particles.
39.997 g/mol is the molar mass of NaOH from adding the masses of Na, O and H. So in theory to create a 1M solution you need to have 39.997 g of Naoh dissolved in 1 dm3 (1 liter).Using the formula Moles = Concentration x volume. you can determine the amount required by plugging in the right data.so you have the concentration and the volume , right? 1.0M and 100mL. you need to change the 100mL to liters. so 100ml = 0.1 liters….So moles required = 0.1 liters x 1.0M. This equals 0.1 moles.From the first line we know that 39.997 grams is equal to 1 moles. Since we need 0.1 moles we can multiply 39.997 to get 3.9997 grams.You then weigh out 3.9997 grams and place it in the volumetric flask, then you fill up to the 100mL line with NaOH.Done!