M=#mol solute / Liter of solution
#mol solute= mass / molar mass
#mol solute= 100 / 87 = 1.15
4= 1.15 / L solution
1.15 / 4 = 0.28
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Since we were only given the (M) and grams of LiBr we had to find the mol of solute. To find the #mol of solute we divided the grams (which were given) over the molar mass of LiBr (87). We then got 1.15 . Now we are able to find the Liter of solution . First we plugged in our given values M= 4 & #mol solute= 1.15 . To find liter of solution we simply dived 1.15 over 4 which equals 0.28.
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I suppose that lithium bromide is not so soluble to prepare 4 M solutions in water at 20 0C.
1.3M
Need moles sodium bromide first. 18.7 grams NaBr (1 mole NaBr/102.89 grams) = 0.1817 moles NaBr =====================Now, Molarity = moles of solute/Liters of solution 0.256 M NaBr = 0.1817 moles NaBr/X Liters Liters = 0.1817/0.256 = 0.7098 Liters -------------------------( you do sigi figis )
The formula for lithium bromide is LiBr. The compound has a molar mass of 86.845 grams per mole. One of its main uses is as a desiccant.
Because you have 6.68 moles of Li2SO4 and 2.500 liters of water, the overall molarity of your solution is 2.67 M.
3.61g/L D=m/v
If 1,1 is grams the molarity is 0,317.
1.3M
Need moles sodium bromide first. 18.7 grams NaBr (1 mole NaBr/102.89 grams) = 0.1817 moles NaBr =====================Now, Molarity = moles of solute/Liters of solution 0.256 M NaBr = 0.1817 moles NaBr/X Liters Liters = 0.1817/0.256 = 0.7098 Liters -------------------------( you do sigi figis )
The formula for lithium bromide is LiBr. The compound has a molar mass of 86.845 grams per mole. One of its main uses is as a desiccant.
.004
Because you have 6.68 moles of Li2SO4 and 2.500 liters of water, the overall molarity of your solution is 2.67 M.
Molarity = moles of solute/Liters of solution ( 250.0 ml = 0.250 liters ) Find moles. 61.7 grams LiCl (1 mole/42.391 grams) = 1.455 moles lithium chloride Molarity = 1.455 moles LiCl/0.250 liters = 5.82 M LiCl -------------------
From the definition of molarity, one liter of the stated solution contains 0.256 mole of sodium bromide. The gram formula weight of sodium bromide is 102.89; therefore, one liter of the solution contains 0.256 X 102.89 or 26.34 grams. 18.7/26.34 = 0.710, the fraction of a liter that contains 18.7 grams of sodium bromide.
To find the molarity of the substance, divide the moles of the substance by the volume of the solution in liters. First, calculate the moles of magnesium bromide: Convert grams to moles by dividing the given mass by the molar mass of magnesium bromide (184.1 g/mol). Moles of magnesium bromide = 4.13 g / 184.1 g/mol = 0.0224 mol. Next, divide the moles by the volume of the solution in liters: Molarity = 0.0224 mol / 845 L ≈ 0.00003 M (rounded to 3 significant figures).
3.61g/L D=m/v
438 grams.
1 percent = 10 grams 2 % = 20 grams x 3 liters = 60 grams