what information about a solution is required to calculate its molarity
Molarity
The content that follows is the substance of lecture 10. In this lecture we comprehend Molarity, Units of Concentration and the Dilution Procedure.
Solution Concentration is an important underlying concept that yous should know well before we outset the next few lectures on Solutions. In other words you tin't understand how to convert and use information regarding solutions unless we first discuss the units in which the concentration of said solutions volition be expressed.
There are 5 main ways we describe the concentration of solutions: 1) Molarity; 2) Molality; 3) Weight Percent; 4) Mole Fraction; and 5) Parts Per Million or Billion. You should know the meaning of each of these terms and more than importantly how to convert from one to the other. I take added a conversion calculator at the lesser of this folio to help yous check your homework etc. But don't get also reliant on it since it will not be available during exams.
Molarity: The molarity of a solution is calculated by taking the moles of solute and dividing past the liters of solution. Molarity is designated by a upper-case letter "Thou".
Molarity = Moles Solute / Liter of Solution
Molality:The molality of a solution is calculated by taking the moles of solute and dividing by the kilograms of solvent. Molality is designated by a lower case "1000". We often express concentrations in molality when we publish because unlike molarity, molality is not temperature dependent. This independence makes it easier for scientists around the earth to reproduce the work. In that location is a simple method for converting molarity to molality:
Here is a uncomplicated example:
Find the molality of 2.v M HiiSOiv. This solution has a density of 1.54 g/mL.
Pace 1. Make an assumption. Assume you have i L of solution. This is a very important step and the amount of solution is not given only y'all need to have a specific quantity to practice the calculations and 1 liter is the best assumption for this problem.
Step 2. Detect the total mass of the solution. Multiply i 50 X the density (1.54 thousand/mL) X 1000 mL/L. This gives 1540 grams of solution.
Step 3. Calculate the grams of the solute. 2.5M means 2.5 moles of sulfuric acrid per i liter of solution. Convert 2.v moles to grams. The molar mass of sulfuric acid is 98.09 g/mol. 2.5 moles X 98.09 g/mol = 245.225 grams of sulfuric acrid.
Step 4. Calculate the grams of the solvent. 1540 grams of solution - 245.225 grams of solute = 1294.775grams or ane.294775 kg solvent
Step five. Summate the molality. 2.5 moles solute / 1.294775 kg solvent = ane.9 molal HtwoThen4
Weight Pct(or Mass Pct): The weight per centum of a solution is calculated by taking the mass of a single solute and dividing it by the total mass of the solution. This can be in grams or kilograms as long as the units of both are expressed in the aforementioned manner. The ratio is so multiplied by 100%. The weight percent is designated by Wt% and sometimes due west/w%.
Mole Fraction: The mole fraction of a single solute in a solution is simply the number of moles of that solute divided past the full moles of all the solutes/solvents. The mole fraction of solutei is written equally Xi.
Parts Per Million(PPM) and Parts per Billion (PPB): "Parts per" is a user-friendly note used for depression and very low concentrations. Generally speaking it is very like to weight per centum - 1% westward/westward means one gram of substance per every 100 1000 of sample and it is (although very rarely) named pph - parts per hundred. Other abbreviations represent:
| ppm | parts per milion (xsix) |
| ppb | parts per bilion (10ix) |
| ppt | parts per trillion (1012) |
| ppq | parts per quadrilion (1015) |
ppq is more than a theoretical construct than a useful measurement, chances are y'all will never see it in apply. Parts per million also tin be expressed as milligrams per liter (mg/50).
For convenience, this worksheet allows you to select dissimilar mass, volume, and concentration units, and the necessary conversions are carried out for you to obtain the value of the blank cell in the desired unit. Notation that the unit of molecular weight must be g/mol.
Serial Dilution:
Serial dilution is a process by which a series of solutions can be made which conserves the amount of chemical needed. The process uses each successively created solution equally the stock solution for the next. The calculation is very simple. The Equation:
The equation allows you to summate how much of the stock solution contains the right amount of moles for the more than dilute solution. Note: Yous can only get from higher to lower concentration. You cannot "concentrate" a solution from lower to higher concentration. The following video runs through the calculations in more detail.
Do Issues:
Molarity Calculations
Serial Dilutions Worksheet
Solutions Practice Problems and Answers
Source: https://chem.fsu.edu/chemlab/chm1045/molarity.html
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