Molarity = moles of solute/volume of solution
1.50 M = X Moles/125ml
= 187.5 millimoles, which is 0.1875 moles
The molar mass of CaCl2 = 110.98 grams
0.1875 moles CaCl2 (110.98g/1mol) = 20.8 grams needed
First, you have to look at the things you already know:
-you are making a calcium chloride solution that is 0.5 M in concentration
-you want 5.71 L of that solution
Next, you have to calculate the formula mass for CaCl2 using the Periodic Table:
Ca weighs about 40 g/mol
Cl weighs about 35.45 g/mol
Since there are 2 chlorines... you need to multiply 35.45 by 2, which gives you 70.9 g/mol. Now add the Ca... so.... 70.9 + 40 = 110.9 g/mol.
Now you have another vital piece of information. The calculation above tells you that each mol of CaCl2 weighs 110.9 grams. Moving on..
The concentration you want to end up with is 0.5 M. 0.5 M is the same as 0.5 mol/L. You also know how many liters you want to make, so you multiply 0.5 mol/L by 5.71 L. This will cancel out the liters and leave you with 2.855 moles.
Now, you just figured out how many MOLES of CaCl2 you will need and you have one last step to figure out how many grams you want. This is where the formula mass comes into play. You have to multiply the number of moles you just calculated by the formula mass. So take 2.855 mol times 110.9 g/mol. And if you pay attention, you'll see that the mol unit cancels out and you're left with 316.6 grams.
CaCl2 Molar mass is 111.0 (to 4 significant figures)
150 ml (0.15L) of 2.0 M solution contains 0.15x2.0 = 0.3 moles
This is 0.3x111.0 = 33.3g of CaCl2 (if it is anhydrous).
1.1 Mols of CaCl2
125 g CaCl2 contain 79,54 g Cl.
Molarity = moles of solute/Liters of solution 0.320 M CaCl2 = moles CaCl2/4.5 Liters = 1.44 moles of CaCl2 1.44 moles CaCl2 (110.978 grams/ 1 mole CaCl2) = 159.81 grams needed so, considering the sigi figis, 160 grams needed.
The molarity is 2,973.
The needed mass is 35,549 g.
The answer is 6,71 g dried KCl.
30 grams
Molarity = moles of solute/Liters of solution 0.320 M CaCl2 = moles CaCl2/4.5 Liters = 1.44 moles of CaCl2 1.44 moles CaCl2 (110.978 grams/ 1 mole CaCl2) = 159.81 grams needed so, considering the sigi figis, 160 grams needed.
The gram formula mass of CaCl2 is 110.99. By definition, each liter of 0.700 M CaCl2 contains 0.700 gram formula masses of the solute. Therefore, 2.00 liters of such solution contain 1.400 formula masses of the solute, or 155 grams, to the justified number of significant digits.
The molarity is 2,973.
1.17 grams :)
The needed mass is 35,549 g.
See the two Related Questions to the left for the answer.The first is how to prepare a solution starting with a solid substance (and dissolving it). The second question is how to prepare a solution by diluting another solution.
The answer is 6,71 g dried KCl.
For this you need the atomic (molecular) mass of CaCl2. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. CaCl2=111.1 grams7.5 grams CaCl2 / (111.1 grams) = .0675 moles CaCl2
600 mL of 0,9 % sodium chloride: 6 x 0,9 = 5,4 grams NaCl
Molarity = moles of solute/Liters of solution 0.75 M KCl = moles KCl/2.25 Liters = 1.6875 moles KCl (74.55 grams/1 mole KCl) = 126 grams of KCl needed
30 grams
160 g (solution) - [5/100*160] g (solute) = 152 g (solvent) water