#### • Class 11 Physics Demo

Explore Related Concepts

# alcohol dilution calculator

Question:If you have a mixed drink....or wine....and you put water in it...it has the same alcohol content right? I know somebody who adds water to drinks and says "dont' worry it's diluted" as though it has no effect anymore. What's up?

Answers:A shot of booze is a shot of booze regardless of what you add; whether it be water, ice that melts, juice etc. It just fills you stomach more.

Question:I started with 1g of a herb having its metal content analysed by AAS. Now, I have multiplied the result that I got by the dilution factor to give me xmg/L or ppm. But I need to determine the amount of metal present in the 1g of herb, how would I go about doing this? I seriously cannot wrap my mind around this..

Answers:Did you extract the entire 1 g of herb??? If so then all of the metal from the 1 g of herb is in the original volume of solution that you started with. So apply your x mg/L to the volume of your extract. eg you had 50.0 mL of extract, all the metal from 1 g is in this 50.0 mL. It has conc. of 0.2 mg/L 50.0 mL = 0.0500 L So amount in 1 g = 0.2 mg/L x 0.0500 L = 0.01 mg in 1 g herb = 0.01 mg/g

Question:I have this biology experiment in which I have to test the amount of diffusion occurring into a decalcified egg (an egg soaked in vinegar that has had its shell disintegrated). Different students chose different liquids, and I happen to have isopropyl alcohol. This has a concentration of 91%, and I have two bottles of about 473 mL (16 fl oz). I need to know how add water to change the concentration of the alcohol (bring it down to 70%, 50% 30%, 10%, that kind of stuff). Can anyone help?

Answers:An emulsifier (also known as an emulgent) is a substance which stabilizes an emulsion by increasing their kinetic stability. One class of emulsifiers are known as surface active substances, or surfactants. Examples of food emulsifiers are egg yolk (where the main emulsifying chemical is lecithin), honey, and mustard, where a variety of chemicals in the mucilage surrounding the seed hull act as emulsifiers; proteins and low-molecular weight emulsifiers are common as well. Soy lecithin is another emulsifier and thickener. In some cases, particles can stabilize emulsions as well through a mechanism called Pickering stabilization. Both mayonnaise and Hollandaise sauce are oil-in-water emulsions that are stabilized with egg yolk lecithin. Detergents are another class of surfactant, and will physically interact with both oil and water, thus stabilizing the interface between oil or water droplets in suspension. This principle is exploited in soap to remove grease for the purpose of cleaning. A wide variety of emulsifiers are used in pharmacy to prepare emulsions such as creams and lotions. Common examples include emulsifying wax, cetearyl alcohol, polysorbate 20, and ceteareth 20.[9] Sometimes the inner phase itself can act as an emulsifier, and the result is nanoemulsion - the inner state disperses into nano-size droplets within the outer phase. A well-known example of this phenomenon, the ouzo effect, happens when water is poured in a strong alcoholic anise-based beverage, such as ouzo, pastis, arak or raki. The anisolic compounds, which are soluble in ethanol, now form nano-sized droplets and emulgate within the water. The colour of such diluted drink is opaque and milky.

Question:Approximately 2.29 mL of concentrated sulfuric acid (specific gravity is 1.84 is added to water. The dilute sulfuric acid is titrated against NaOH and the concentration of H2SO4 is 0.184 mol dm^-3. How do I calculate the molality of the acid produced by the dilution?

Answers:Essentially, you're given the molarity of the dilute acid, from which you can deduce the molality by finding the weight of the water in 1 L of solution. Step 1: Find out the mass of acid in 1 L of solution. mass = .184 moles * molecular mass of H2SO4 2 H (1) + 4 O (16) + S (32) = 96 g/mol .184 mol * 96 g/mol = 17.664 g H2SO4 / L Step 2: Using the mass and specific gravity, find the volume of acid. 17.664 g / 1.84 g/cm^3 = 9.239 cm^3 of acid Step 3: Given the volume of acid, and 1 liter of solution, find the volume of water. Since the density of water is 1 g/cm^3, that will also equal the mass in g. Vw = 1000 cm^3 (1 L) solution - 9.239 cm^3 of acid Vw = 990.761 cm^3 mw = 990.761 g Step 4: Find molality given # of moles of H2SO4 and weight of the water in 1 L of solution: molality = .184 mol / .990761 kg water molality = 0.182 molal (solution!)