The molality of 2.5 g of ethanoic acid (CH3COOH) in 25g of benzene is 

  • Option 1)

    0.167m

  • Option 2)

    1.67m

  • Option 3)

    0.556m

  • Option 4)

    0.00167m

 

Answers (1)
P Plabita

As we learned in concept

Molality -

Molality = \frac{Moles \: of \: solute}{Mass \, \: o\! f\: solution(Kg)}

-

 

 moles of ethanoic acid =  \frac{2.5}{60}

Mass of solvent =25g=0.025 kg

Molality , m=\frac{2.5}{60\times0.025}

=1.67m

 


Option 1)

0.167m

Option is incorrect

Option 2)

1.67m

Option is correct

Option 3)

0.556m

Option is incorrect

Option 4)

0.00167m

Option is incorrect

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