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How many mL are required to make a 2.50 M solution with 385.5g of BaCl2?
Wiki User
∙ 15y ago
use the equation:
(formula weight in g/mol)(Molarity needed in mol/L)(volume required in L)= amt reagent needed (in grams)
Make sure to keep all of the units and make sure they canell out.
(F.W. of BaCl2 in g/mol)(2.50 mol/L)(x in Liters) = 385.5 g
After you do the calculation, you will need to convert the answer from Liters to Milliliters by :
x in L * (1000 mL / 1 L)
Wiki User
∙ 15y ago
Wiki User
∙ 11y agoStep 1: get the molar mass of magnesium chloride. Mg = 24 and Cl = 35.5, so 24+35.5+35.5=95g/mol.
Step 2: A 2.5M solution is 2.5 moles per liter, so 95x2.5=237.5g MgCl2/liter.
Step 3: You only need 500mL (0.5L) of it, so 237.5/2=118.75g MgCl2.
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