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How many moles of solute are present in 1.5 liters of 1.33 M of Na2SO4?
Wiki User
∙ 8y ago
The answer is approx. 2 moles (for anhydrous sodium sulfate).
Wiki User
∙ 8y ago
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How many moles of HCl are present in 40ml of a 0.035M solution?
Molarity = moles of solute/Liters of solution (40 ml = 0.04 Liters) algebraically manipulated, Moles of solute = Liters of solution * Molarity Moles HCl = (0.04 Liters)(0.035 M) = 0.0014 moles HCl ==============
How many moles of sodium chhloride are present in 22.0 ml of 0.500 m solution?
Molarity = moles of solute/Liters of solution ( 22.0 ml = 0.022 Liters ) moles of solute = Molarity * Liters of solution Moles of NaCl = 0.500 M * 0.022 Liters = 0.011 moles of sodium chloride -------------------------------------------
How many moles of MgCl2 are present in 20.0 mL of 0.800 M MgCl2 solution?
Molarity = moles of solute/Liters of solution ( 20.0 ml = 0.02 Liters ) moles of solute = Liters of solution * Molarity 0.02 Liters * 0.800 M MgCl2 = 0.016 moles MgCl2 -------------------------------
How many moles of Na2CO3 are there in 10.0 L of 2.0 M soluton?
Molarity = moles of solute/Liters of solution Or, for our purposes, Moles of solute = Liters of solution * Molarity Moles Na2CO3 = 10.0 Liters * 2.0 M = 20 moles Na2CO3 --------------------------
How many moles of KMnO4 are present in 75.0 mL of a 0.0950 M solution?
Molarity = moles of solute/Liters of solution ( 75.0 ml = 0.075 Liters ) Algebraically manipulate, moles of solute = Liters of solution * Molarity Moles KMnO4 = (0.075 Liters)(0.0950 M) = 7.13 X 10 -3 moles KMnO4 ------------------------------------
What is the concentration of Na2SO4 solution prepared by dissolving 5.35 g of Na2SO4 in sufficient water to give 330 ml of solution?
First.Get moles sodium sulfate.5.35 grams Na2SO4 (1 mole Na2SO4/142.05 grams)= 0.0377 moles Na2SO4-------------------------------------Second.Molarity = moles of solute/Liters of solution ( 330 mL = 0.33 Liters )Molarity = 0.0377 moles Na2SO4/0.33 Liters= 0.114 M Na2SO4=============
What is the molarity of sodium sulphate in a solution prepared by dissolving 15.5g in enough water to make 35ml of solution?
Molarity = moles of solute/volume of solution Find moles NaCl 55 grams NaCl (1mol NaCl/58.44 grams) = 0.941 moles NaCl Molarity = 0.941 moles NaCl/35 Liters = 0.027 Molarity NaCl ( sounds reasonable as 55 grams is not much in 35 Liters of water, which would be about 17.5 2 liter sodas )
How many moles of HCl are present in 40ml of a 0.035M solution?
Molarity = moles of solute/Liters of solution (40 ml = 0.04 Liters) algebraically manipulated, Moles of solute = Liters of solution * Molarity Moles HCl = (0.04 Liters)(0.035 M) = 0.0014 moles HCl ==============
How many moles of sodium chhloride are present in 22.0 ml of 0.500 m solution?
Molarity = moles of solute/Liters of solution ( 22.0 ml = 0.022 Liters ) moles of solute = Molarity * Liters of solution Moles of NaCl = 0.500 M * 0.022 Liters = 0.011 moles of sodium chloride -------------------------------------------
How many moles of MgCl2 are present in 20.0 mL of 0.800 M MgCl2 solution?
Molarity = moles of solute/Liters of solution ( 20.0 ml = 0.02 Liters ) moles of solute = Liters of solution * Molarity 0.02 Liters * 0.800 M MgCl2 = 0.016 moles MgCl2 -------------------------------
How many moles of KCL are present in 50.0ml of a 0.552 M solution?
Molarity = moles of solute/Liters of solution ( 50.0 ml = 0.05 Liters ) 0.552 M KCl = moles/.0.05 liters = 0.0276 moles of KCl
How many moles of Na2CO3 are there in 10.0 L of 2.0 M soluton?
Molarity = moles of solute/Liters of solution Or, for our purposes, Moles of solute = Liters of solution * Molarity Moles Na2CO3 = 10.0 Liters * 2.0 M = 20 moles Na2CO3 --------------------------
How many moles of KMnO4 are present in 75.0 mL of a 0.0950 M solution?
Molarity = moles of solute/Liters of solution ( 75.0 ml = 0.075 Liters ) Algebraically manipulate, moles of solute = Liters of solution * Molarity Moles KMnO4 = (0.075 Liters)(0.0950 M) = 7.13 X 10 -3 moles KMnO4 ------------------------------------
How many moles of kbr are present in 25ml of a 1.5 M solution?
Molarity = moles of solute/Liters of solution ( 25 ml = 0.025 Liters ) 1.5 M KBr = moles KBr/0.025 Liters = 0.038 moles potassium bromide ------------------
How many moles of sodium ions are present in 2.50 L of 0.300 M Na3PO4?
Molarity = moles of solute/Liters of solution 0.300 M Na3PO4 = moles Na3PO4/2.50 Liters = 0.75 moles Na3PO4
What is the molarity a 500.0 ml solution containing 0.75 moles of solute?
Molarity = moles of solute/Liters of solution Without the solute name the mass ( 8 grams ) does no good. Mass of solute (1 mole/molar mass of solute) = moles solute ----------------------then use Molarity equation. ( remember convert to liters )
How many grams of solid barium sulfate form when 22.6 mL of 0.160 M barium chloride reacts with 54.6 mL of 0.055 M sodium sulfate?
Balanced equation first. BaCl2 + Na2SO4 -> 2NaCl + BaSO4 22.6 ml BaCl2 = 0.0226 liters 54.6 ml Na2SO4 = 0.0546 liters 0.160 M BaCl2 = moles BaCl2/0.0226 liters = 0.00362 moles BaCl2 0.055 M Na2SO4 = moles Na2SO4/0.0546 liters = 0.0030 moles Na2SO4 The ratio of BaCl2 to Na2SO4 is one to one, so either mole count wull drive this reaction. Use 0.0003 moles Na2SO4 0.0030 moles Na2SO4 (1 mole BaSO4/1 mole Na2SO4)(233.37 grams/1 mole BaSO4) = 0.700 grams of BaCO4 produced
