Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

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)

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

Posted by

Vakul

View full answer

Similar Questions

Latest Question