Molarity Help
Determine the Molarity (M, moles solute / Liter of solution) of a solution formed when 1.734E6 µg Barium hydroxide is dissolved in water such that the final volume of the solution is 39.56 cL.
Determine the Molarity (M, moles solute / Liter of solution) of a solution formed when 1.534 g Cobalt(II) hydrogen sulfate is dissolved in water such that the final volume of the solution is 375.6 mL.
Determine the Molarity (M, moles solute / Liter of solution) of a solution formed when 4.711 µg Sodium hypochlorite is dissolved in water such that the final volume of the solution is 2.000 mL
Determine the Molarity (M, moles solute / Liter of solution) of a solution formed when 6.656E22 molecules of Tin(IV) chlorate are dissolved in water such that the final volume of the solution is 10.90 cL
For conversion use below equation:
1 µg = 1 g
1 cL = 10 mL = 0.01 L
Given that,
Mass of Barium hydroxide = 1.734E6 µg
= 1.734 µg
= 1.734
= 1.734 g
Volume of solution = 39.56 cL
= 39.56 0.01
= 0.3956 L
No: of moles of barium hydroxide in 1.734 g = Mass / Molar mass
= 1.734g / 171.341g/mol
= 0.01012 mol
Therefore, molarity of barium hydroxide solution = no: of moles / volume of solution in L
= 0.01012 mol/0.3956 L
= 0.02558 mol/L
= 0.02558 M
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Mass of Cobalt(II) hydrogen sulfate [Co(HSO4)2] = 1.534 g
Volume of solution = 375.6 mL
= 375.6
= 0. 3756 L
Molar mass of Cobalt(II) hydrogen sulfate [Co(HSO4)2] = 253.0743g/mol
No: of moles of Cobalt(II) hydrogen sulfate in 1.534 g = Mass / Molar mass
= 1.534g / 253.0743g/mol
= 6.06146 mol
= 0.00606146 mol
Therefore, molarity of Cobalt(II) hydrogen sulfate solution= no: of moles / volume of solution in L
= 0.00606146 mol /0. 3756 L
= 0.016138 M
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Mass of Sodium hypochlorite [NaClO] = 4.711 µg
= 4.711 µg
= 4.711 g
Volume of solution = 2.000 mL
= 2.000
= 0.002 L
Molar mass of Sodium hypochlorite [NaClO] = 22.98977 + 35.453 + 15.9994 = 74.44217 g/mol
No: of moles of Sodium hypochlorite in 4.711 g = Mass / Molar mass
= 4.711 g / 74.44217g/mol
= 6.32840 mol
Therefore, molarity of Sodium hypochlorite Solution= no: of moles / volume of solution in L
= 6.32840 mol / 0.002 L
= 3.164201 M
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1 mol contains avagadros no : of molecules, 6.022 10^23
Therefore, No: of moles in 6.656E22 molecules of Tin(IV) chlorate = 6.656 10^22 / 6.022 10^23
= 0.110528 moles
Molar mass of Tin(IV) chlorate [Sn(ClO3)4] = 452.5140 g/mol
Volume of solution = 10.90 cL
= 10.90 0.01
= 0.1090 L
Therefore, molarity of Tin(IV) chlorate solution = no: of moles / volume of solution in L
= 0.110528 mol / 0.1090 L
= 1.01401 M
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