Wiki User
∙ 11y agoOne percent by weight.
Alcohol has a density of less than water, so 1 g of alcohol is more than 1 mL.
Wiki User
∙ 11y agoSuppose there is x% by volume of something in a mixture then it simply means out of 100 ml of that sample x ml is of that thing. ------------------------------------------------------------------- There are two common ways of indicating how much of each element or compound is contained in a mixture. They are percent by volume and percent by weight. For example, the amount of alcohol in an alcoholic beverage is measured in proof, where one proof equals one half percent alcohol by volume.
Percent of an objects mass is expressed in terms of its weight. Percent of an objects volume is expressed in terms of its size.
If the percents given are by weight or mass, this is very straightforward: The ratio between the desired percentage and the initial percentage is 1/50. Therefore, a given mass of initial solution must be diluted to 50 times its original mass to obtain the desired lower concentration, or in other words, 49 parts of diluent must be mixed with each part of initial solution. If the percents involve volume measurements, it would be necessary to take into account and change in density occasioned by the dilution.
That depends on the volume of the brick. Whatever its volume is, its weight underwater is(weight of the brick in air) minus (weight of an equal volume of water)
Bulk density = dry weight / volume, then by knowing the dry weight and bulk density we can calculate the volume.
(Note: This answer assumes that the "ounces" specified are avoirdupois or other weight ounces and that percentages are by weight; otherwise possible volume changes on dilution must by considered.) The weight of pure alcohol in each solution is the product of the percentage and the total weight of the solution. Therefore, designating the unknown weight of 30 % alcohol as w, from the problem statement 0.30w + 0.80(40) = 0.70(w + 40), or 0.30w + 32 = 0.70w + 28, or 32 - 28 = w(0.70 - 0.30) or w = 4/0.40 = 10 ounces of 30 % alcohol.
It means that either 86% of the weight is alcohol, or 86% of the volume. Volume is more standard, I believe.It means that either 86% of the weight is alcohol, or 86% of the volume. Volume is more standard, I believe.It means that either 86% of the weight is alcohol, or 86% of the volume. Volume is more standard, I believe.It means that either 86% of the weight is alcohol, or 86% of the volume. Volume is more standard, I believe.
To make a percent sucrose solution, dissolve a specific weight of sucrose in a specific volume of water. For example, to make a 10% sucrose solution, dissolve 10 grams of sucrose in 90 mL of water. The formula to calculate the amount of sucrose needed is: (percent sucrose/100) x volume of solution = weight of sucrose (in grams).
A 1% solution normally contains 1 gram of active ingredient per 100 ml of solution (weight-volume percent) Could also be 1gm per 100 gms (weight-weight percent)- but normally weight-volume is used.
The weight of 10 percent acetic acid solution would depend on the total volume of the solution. For example, if you have 100 grams of a 10 percent acetic acid solution, it would contain 10 grams of acetic acid.
400 mls would require 40g of glucose for a 10% solution and thus 20g for a 5% solution.
About 80ml of water must be added to 40ml of a 25 percent by weight solution to make a 2 percent by weight solution.
You can determine the number of grams of an active ingredient in a solution by multiplying the percent strength of the solution by the total weight or volume of the solution. This will give you the weight of the active ingredient present in the solution.
Suppose there is x% by volume of something in a mixture then it simply means out of 100 ml of that sample x ml is of that thing. ------------------------------------------------------------------- There are two common ways of indicating how much of each element or compound is contained in a mixture. They are percent by volume and percent by weight. For example, the amount of alcohol in an alcoholic beverage is measured in proof, where one proof equals one half percent alcohol by volume.
% volume
To prepare a 1% phenolphthalein solution in 2000mL, you would need to dissolve 20g of phenolphthalein in 2000mL of solvent (usually water or alcohol). This will result in a 1% solution by weight/volume. Remember to wear appropriate personal protective equipment and handle chemicals with care.
To find the percent by weight of CH3CO2H in vinegar, you need to know the molar mass of CH3CO2H and the density of vinegar. Once you have that information, you can calculate the weight of CH3CO2H in a given volume of vinegar, and then express that weight as a percentage of the total weight of the solution.