What is the concentration (in M) of the chloride on when 19.2 mL of a 0.664 M so

Question

What is the concentration (in M) of the chloride on when 19.2 mL of a 0.664 M solution of barium chloride is combined with 18.9 mL of a 0.68 M solution of aluminum chloride? Assume the volumes are additive. What is the concentration of a solution (in M) which contains 56.9 g of aluminum nitrate in 526 mL of solution? What is the concentration of the ammonium ion (in M) in a solution which contains 50.4 g ammonium sulfide in 621 mL of solution? What is the concentration of the sodium ion (in M) in a solution of 0.513 M sodium carbonate? What is the final concentration (in M) of 16.5 mL iron(III) chloride solution with an initial concentration of 1.77 M which is diluted with 18.2 mL of solvent? Assume the volumes are additive. What volume (in mL) of 1.78 M solution of sodium chloride is needed to make 59.6 mL of a 0.621 M solution? Assume the volumes are additive. The units on molarity are A. moles of solute per liter of solution. B. moles of solute per liter of solvent. C. moles of solvent per liter of solution. D. moles of solvent per liter of solute. E. mass of solute per Mar of solution. F. mass of solute per liter of solvent G. mass of solvent per liter of solute. H. mass of solvent per liter of solution. When diluting a solution by adding solvent, what is constant? A. moles of solute B. moles of solvent C. volume of solvent D. volume of solution what is the experimental yield (in g of precipitate) when 15.3 mL of a 0.724 M solution of barium hydroxide is combined with 18.1 mL of a 0.714 M solution of aluminum nitrate at a 81.1% yield?

Explanation / Answer

Resulting molarity M = M1V1+M2V2/V1+V2
M = ((0.664*19.2)+(18.9*0.68))/(19.2+18.9)
M = 0.6719 mol/L

Molarity = mass/molar mass *1000/volume in mL
M = 56.9/213*1000/526
M = 0.5079 mol/L

M = 50.4/50.11*1000/621
M = 1.6196 mol/L

Na2CO3 ------> 2Na+ + CO3 2-
so 1 mole of Na2CO3 produces 2 moles of Na+ ions then
Concentration of Na+ ions = 2*0.513 = 1.026 mol/L

M1V1 = M2V2
1.77*16.5 = M2 *(16.5+18.2)
M2 = 0.8416 mol/L

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