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This question provides us –

  • Weight of [tex] \bf KCl [/tex] is = 47 g
  • Volume, V = 375 mL

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  • Molar Mass of [tex]\bf KCl [/tex] –

[tex]\qquad[/tex] [tex]\twoheadrightarrow\bf 39.0983 \times 35.453 [/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\bf 74.5513[/tex]

Using formula –

[tex]\qquad[/tex] [tex]\purple{\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{ W\times 1000}{MV}}[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{ 47 \times 1000}{74.5513\times 375}[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{47000}{27956.7375}[/tex]

[tex]\qquad[/tex] [tex]\twoheadrightarrow\bf Molarity _{(Solution)} = \cancel{\dfrac{47000}{27956.7375}}[/tex]

[tex]\qquad[/tex] [tex] \twoheadrightarrow\bf Molarity _{(Solution)} = 1.68117M [/tex]

[tex]\qquad[/tex] [tex]\pink{\twoheadrightarrow\bf Molarity _{(Solution)} = 1.7M} [/tex]

  • Henceforth, Molarity of the solution is = 1.7M

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