* First! Figure out what pyridine's formula is: C5H5N So, we know that each molecule of pyridine has 5 C, 5 H, and 1 N. The cool thing about the formula is it doesn't matter what the unit is, as long as the ratio between the C, H, and N remains the same, BUT, being that we're doing chemistry let's use moles. :) 5 moles of C + 5 moles of H + 1 mole of N -->1 mole of C5H5N This can be a weird concept, because you might think "how do you put in 11 moles of stuff and get only one mole of product"... but, just like baking, you put in 5 cups of sugar, 5 cups of flour, and 1 cup of butter, but get one really big cookie (didn't say it would be a good cookie). Well, you put in a few moles of this, a few moles of that...and you get a single mole of C5H5N. * After you figure out the chemical formula, calculate molecular mass: 5 moles of carbon = 5 x 12.011 grams per mole = 60.055 grams 5 moles of hyrdrogen = 5 x 1.008 grams per mole= 5.04 grams 1 mole of nitrogen = 1 x 14.007 grams per mole= 14.007 grams ADD: 60.055 + 5.04 + 14.007 = 79.102 grams per 1 mole of C5H5N * Convert solute to moles! Now, we've figured out that 1 mole of C5H5N has a mass of 79.102 grams....but, we're only putting 5 grams in the solution. How many moles is 5 grams if 1 mole is 79.102 grams? Calculate: 5 grams/ 79.102 grams = .063 moles * Convert to moles per liter (not moles to part of a liter)! So...we are adding .063 moles to .5 liters of solvent, but, molarity is always expressed in moles per liter, so we have to convert this: (.063 moles C5H5N/.5 liter) x 1 liter = .126 moles C5H5N So, if .063 moles of C5H5N dissolved in .5 liters gives a solution of the same concentration as 1.26 moles of C5H5N dissolved in 1 liter (note - we now have it in moles per liter!). As molarity is moles per liter...and adding 5 grams of pyridine (which is .063 moles) to .5 liters of solvent produces a solution of the same concentration (molarity) as dissolving 1.26 moles of solvent in 1 liter, we have produced a 1.26 molar solution.
2.5m
Molarity = moles of solute/Liters of solution Molarity = 5 moles solute/4.5 Liters of solution = 1 M solution ==========
Molarity = moles of solute/Liters of solution
Molarity = moles of solute/Liters of solution Without the solute name the mass ( 8 grams ) does no good. Mass of solute (1 mole/molar mass of solute) = moles solute ----------------------then use Molarity equation. ( remember convert to liters )
Some conversion required. (mmolar into mol, or moles into mmol ) Molarity = moles of solute/Liters of solution 100 millimolar = 0.1 M glycine Molarity = moles of solute/Liters of solution manipulate algebraically Liters of solution = moles of solute/Molarity 0.005 mole glycine/0.1 M glycine = 0.05 Liters ( 1000 ml/1 L) = 50 milliliters of solution --------------------------------
Molarity is defined as moles of solute per liter of solution. In this case, moles of solute is 0.02 and liters of solution is 0.615 (615 mls x 1 L/1000ml). Thus molarity = 0.02/0.615 = 0.0325 M = 0.03 M to 1 significant figure.
Molarity = moles of solute/Liters of solution Molarity = 5 moles solute/4.5 Liters of solution = 1 M solution ==========
Molarity = moles of solute/Liters of solution Molarity = 0.202 moles KCl/7.98 Liters = 0.253 M KCl solution ================
The molarity of a solution containing 7.0 moles of solute in 569mL of solution is a)81M b)0.081M c)12M D)4.0M The answer is C. 12M 7 x 1000/569 = 12.302
31 grams HCl (1mol/36.458g ) = 0.850 moles 500cm^3 = 0.50 Liters Molarity = moles of solute/liters of solution Molarity = 0.850mol/0.50L = 1.7 Molarity
Molarity = moles solute/ Liters solution
Molarity = moles of solute/Liters of solution
The concentration in moles of a substance in the solution
Molarity = moles of solute/Liters of solution Without the solute name the mass ( 8 grams ) does no good. Mass of solute (1 mole/molar mass of solute) = moles solute ----------------------then use Molarity equation. ( remember convert to liters )
Some conversion required. (mmolar into mol, or moles into mmol ) Molarity = moles of solute/Liters of solution 100 millimolar = 0.1 M glycine Molarity = moles of solute/Liters of solution manipulate algebraically Liters of solution = moles of solute/Molarity 0.005 mole glycine/0.1 M glycine = 0.05 Liters ( 1000 ml/1 L) = 50 milliliters of solution --------------------------------
Molarity = moles of solute/liters of solution Molarity = 0.597 moles HCl/0.169 liters = 3.53 M HCl ------------------
Molarity.
Molarity = moles of solute/Liters of solution Molarity = 10 moles salt/20 Liters solution = 0.50 M salt solution ----------------------------