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.
1.17 grams :)
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
8.9g
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.
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
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
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