93,31 g MgCl2 are needed.
Need moles MgCl2 75.0 grams MgCl2 (1 mole MgCl2/95.21 grams) = 0.7877 mole MgCl2 ================now, Molarity = moles of solute/Liters of solution ( 500.0 milliliters = 0.5 Liters ) Molarity = 0.7877 moles MgCl2/0.5 Liters = 1.58 M MgCl2 solution --------------------------------
MgCl2 Molarity = moles of solute/Liters of solution ( 250 ml = 0.250 L ) Get moles MgCl2 80 grams MgCl2 (1 mole MgCl2/95.21 grams) = 0.8402 moles MgCl2 Molarity = 0.8402 moles MgCl2/0.250 Liters = 3.4 M MgCl2 ----------------
7.5 grams are needed because each milliliter is 1.50 grams, to get 5.00 mls, the calculation would be 5x1.50=7.5.
HCl has a molar mass of 36.461 grams per mole. This means that 72.922 grams of HCl are needed per liter of water to make a solution that has a concentration of 2M.
Molarity= Number of moles of solute/Liters of solution 50 grams KOH 700 ML to .7 Liters of h2o Molar Mass of KOH= 56 50 divided by 56 = .89 moles Molarity= .89 mol/.7 L = 1.27 MOLARITY
Need moles MgCl2 75.0 grams MgCl2 (1 mole MgCl2/95.21 grams) = 0.7877 mole MgCl2 ================now, Molarity = moles of solute/Liters of solution ( 500.0 milliliters = 0.5 Liters ) Molarity = 0.7877 moles MgCl2/0.5 Liters = 1.58 M MgCl2 solution --------------------------------
MgCl2 Molarity = moles of solute/Liters of solution ( 250 ml = 0.250 L ) Get moles MgCl2 80 grams MgCl2 (1 mole MgCl2/95.21 grams) = 0.8402 moles MgCl2 Molarity = 0.8402 moles MgCl2/0.250 Liters = 3.4 M MgCl2 ----------------
The molar mass of MgCl2 is 95.21 g. Therefore 0.96 grams is 0.01008 mol of the substance. So the molarity of the solution is 5.04 x 10-3 M.
Molarity = moles of solute/volume of solution. 0.73 grams MgCl2 (1mol/95.21g ) = 0.00767 moles/0.300 Liters ( to get M, not mM ) = 2.6 X 10^-2 M
1 mole of MgCl2=95.21 grams (found on the periodic table) 17.5 g of MgCl2(1mole of MgCl2/95.21g of MgCl2)=.184 moles of MgCl2(grams cancel out and rounded to 3 sig. figs) .184 moles of MgCl2 (1 Liter/2.5 moles of MgCl2)=0.735L of MgCl2 (moles cancel and rounded to 3 sig figs) 0.735 L of MgCl2(1000 milliLiters/1 Liter)=73.5 mL of MgCl2 and round the answer to 2 sig figs which equals 74 milliLitersof MgCl2
Magnesium chloride (MgCl2) has a molecular weight of 95.21 grams per mole. 11.6 grams of MgCl2 is therefore .122 moles.
7.5 grams are needed because each milliliter is 1.50 grams, to get 5.00 mls, the calculation would be 5x1.50=7.5.
HCl has a molar mass of 36.461 grams per mole. This means that 72.922 grams of HCl are needed per liter of water to make a solution that has a concentration of 2M.
Calculate the mass (in grams) of sodium sulfide that is needed to make 360ml of a 0.50 mol/L solution
Molarity= Number of moles of solute/Liters of solution 50 grams KOH 700 ML to .7 Liters of h2o Molar Mass of KOH= 56 50 divided by 56 = .89 moles Molarity= .89 mol/.7 L = 1.27 MOLARITY
0.50 grams of BeCl2
45 g MgCl2 x 1 mole MgCl2/95.2 g /0.5 L = 0.9045 moles/L = 0.90 M (to 2 significant figures)