Question

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

Answer 1

• Molarity of solution when 1.734E6 µg Barium hydroxide is dissolved in water such that the final volume of the solution is 39.56 cL

For conversion use below equation:

1 µg = 1 $\times$ $10^{-6}$ g

1 cL = 10 mL = 0.01 L

Given that,

Mass of Barium hydroxide = 1.734E6 µg

= 1.734 $\times$ $^{}$$10^{6}$ µg

= 1.734 $\times$ $^{}$$10^{6}$$\times$$^{}$$10^{-6}$

= 1.734  g

Volume of solution = 39.56 cL

= 39.56 $\times$ 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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• Molarity 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

Mass of Cobalt(II) hydrogen sulfate [Co(HSO4)2] = 1.534 g

Volume of solution = 375.6 mL

= 375.6 $\times$ $10^{-3}$

= 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 $\times$ $10^{-3}$ 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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• Molarity 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

Mass of Sodium hypochlorite [NaClO] = 4.711 µg

= 4.711 µg

= 4.711 $\times$ $^{}$$10^{-6}$ g

Volume of solution = 2.000 mL

= 2.000 $\times$ $10^{-3}$

= 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 $\times$ $^{}$$10^{-6}$ g = Mass / Molar mass

= 4.711 $\times$ $^{}$$10^{-6}$g /  74.44217g/mol

= 6.32840 $\times$ $10^{-8}$ mol

Therefore, molarity of Sodium hypochlorite Solution= no: of moles / volume of solution in L

= 6.32840 $\times$ $10^{-8}$ mol / 0.002 L

= 3.164201 $\times$ $10^{-5}$ M

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• Molarity 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

​​​​​​​​​​​​​​1 mol contains avagadros no : of molecules, 6.022 $\times$ 10^23

Therefore, No: of moles in 6.656E22 molecules of Tin(IV) chlorate = 6.656 $\times$ 10^22 /  6.022 $\times$ 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  $\times$ 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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