How many grams of NaCl are contained in 350. mL of a 0.334 M solution of sodium chloride?
Which of the following aqueous solutions contains the greatest number of ions?
What mass of calcium chloride is needed to prepare 3.650 L of a 1.75 M solution?
A 57.17-g sample of Barium Hydroxide, Ba(OH)2 is dissolved in enough water to make 1.800 liters of solution. How many mL of this solution must be diluted with water in order to make 1.000 L of 0.100 M Ba(OH)2?
An analytical procedure requires a solution of chloride ions. How many grams of calcium chloride must be dissolved to make 2.15 L of 0.0520 M chloride ions?
Which of the following aqueous solutions contains the greatest number of ions?
What mass of calcium chloride is needed to prepare 3.650 L of a 1.75 M solution?
A 57.17-g sample of Barium Hydroxide, Ba(OH)2 is dissolved in enough water to make 1.800 liters of solution. How many mL of this solution must be diluted with water in order to make 1.000 L of 0.100 M Ba(OH)2?
An analytical procedure requires a solution of chloride ions. How many grams of calcium chloride must be dissolved to make 2.15 L of 0.0520 M chloride ions?
-
1st question, .334 M means you have .334mol NaCl / I L of solution, because M (molarity) is moles per liter. so set up a ratio of .334mol/L = x mol/.350 ml (convert mL to L) solve for x and you should get .1204 moles of NaCl in 350 mL of a .334 M solution. Now use the molar mass of NaCl, which you can find looking at your periodic table, it's 58.5 g, about.
Now, multiply the moles you found by the molar mass value. .1204mol NaCl x (58.5g NaCl/ 1 mol NaCl) moles will cancel so you will be left with a value in grams, which is what you're looking for, I got 7.04 g NaCl.
2nd question, You didn't list any 'following aqueous soluitons' so I can't help you there.
3rd question. 1.75 M means 1.75mol per 1L of solution. set up the ratio:
1.75mol CaCl2/1L = xmol CaCl2/3.65L and solve for x. I got 6.3875 mol CaCl2 in 3.65 L of solution.
use the same process as in part one to go from moles to grams.
6.3875 mol CaCl2 x (111g CaCl2/ 1 mol) I got 709 grams.
4th question: You have 57.17 g of Ba(OH)2, convert to moles using molar mass...
57.17g x (1 mol/111g) = .334 mol of Ba(OH)2. set up a ratio similar to the other problems..
.334mol/1.8L = x mol/1L solve for x, I got x = .1856 mol set up another ratio, this time solving for volume,
.1856mol/1L = .100mol/xL solve for x... I got x = .539 L, convert to mL = 539mL. Not 100% sure here, but it seems right.
5th, There are 2 Cl molecules in every atom of CaCl2 so the molarity of Cl will be twice that of CaCl2
or the molarity of CaCl2 is half that of Cl, so you'll be working with .026 moles of solution.
Set up the same type of ratio...
.026/1L = xmol/2.15L, solve for x... x=.0559 mol CaCl2 required, convert to grams...
.0559 x (111g/1mol) = 6.205 grams of CaCl2 needed.
Hope this helps.
Now, multiply the moles you found by the molar mass value. .1204mol NaCl x (58.5g NaCl/ 1 mol NaCl) moles will cancel so you will be left with a value in grams, which is what you're looking for, I got 7.04 g NaCl.
2nd question, You didn't list any 'following aqueous soluitons' so I can't help you there.
3rd question. 1.75 M means 1.75mol per 1L of solution. set up the ratio:
1.75mol CaCl2/1L = xmol CaCl2/3.65L and solve for x. I got 6.3875 mol CaCl2 in 3.65 L of solution.
use the same process as in part one to go from moles to grams.
6.3875 mol CaCl2 x (111g CaCl2/ 1 mol) I got 709 grams.
4th question: You have 57.17 g of Ba(OH)2, convert to moles using molar mass...
57.17g x (1 mol/111g) = .334 mol of Ba(OH)2. set up a ratio similar to the other problems..
.334mol/1.8L = x mol/1L solve for x, I got x = .1856 mol set up another ratio, this time solving for volume,
.1856mol/1L = .100mol/xL solve for x... I got x = .539 L, convert to mL = 539mL. Not 100% sure here, but it seems right.
5th, There are 2 Cl molecules in every atom of CaCl2 so the molarity of Cl will be twice that of CaCl2
or the molarity of CaCl2 is half that of Cl, so you'll be working with .026 moles of solution.
Set up the same type of ratio...
.026/1L = xmol/2.15L, solve for x... x=.0559 mol CaCl2 required, convert to grams...
.0559 x (111g/1mol) = 6.205 grams of CaCl2 needed.
Hope this helps.
-
For the first question, you need to find the moles. The formula for moles is n = c x v where 350 mL is the v and 0.334 is the concentration (or molarity denoted by M). So 0.350 (because you have to convert to liters) times 0.334 is 0.1169 moles. Formula for mass: mass = moles times molar mass.