The molalility of a 15 % (w/v) solution of H_{2}SO_{4} of density of 1.1 g/cm^{3} approx is

  • Option 1)

    1.2

  • Option 2)

    1.4

  • Option 3)

    1.8

  • Option 4)

    1.6

 

Answers (1)
V Vakul

As we learned in concept

Molality -

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

-

 

 15g\: of\: H_{2}SO_{4}\rightarrow 100 ml\: of\: solution

=100\times1.1 g\: of\: solution

=110g\: of\: solution

Molality=\frac{moles}{wt.\ of\ solvent\ in\ kg }

=\frac{\frac{15}{98}\times1000}{(110-15)}

=\frac{15\times1000}{98}\times\frac{1}{95}

=1.61\:m\sim 1.60\:m


Option 1)

1.2

Option is incorrect

Option 2)

1.4

Option is incorrect

Option 3)

1.8

Option is incorrect

Option 4)

1.6

Option is correct

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