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Calculate the following quantity: volume in liters of .315M manganese II sulfate that contains 53.0 g of solute
Update:I keep getting .817 L but apparently that's wrong
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53.0gMnSO4 x 1mol/151g x 1L/0.315 mol = 1.11L
Take the 53.0g and convert to moles using molar mass, then convert to liters using molarity factor (0.315 mol per L)
Convert grams of potassium sulfate to grams of potassium sulfate making use of the formula "moles = grams/molecular weight". Convert 40.00 ml to liters (a million liter = 1000ml) Then use the Molarity formula