# Question b0614

May 8, 2017

${\text{1 mol kg}}^{- 1}$

#### Explanation:

In this case, you're not really interested in the density of the solution because molality is defined as the number of moles of solute present per kilogram of solvent.

Use the density of water to find the mass of the solvent

100 color(red)(cancel(color(black)("mL"))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = "100 g"

Use the molar mass of sulfuric acid to convert its mass to moles

98 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"SO"_4)/(98.08color(red)(cancel(color(black)("g")))) = "0.9992 moles H"_2"SO"_4

The number of moles of sulfuric acid present in $\text{1 kg}$ of solvent will be equal to

1 color(red)(cancel(color(black)("kg solvent"))) * (10^3 color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * ("0.9992 moles H"_2"SO"_4)/(100color(red)(cancel(color(black)("g solvent"))))= "9.992 moles H"_2"SO"_4#

You can thus say that the solution has a molality of

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molality = 1 mol kg}}^{- 1}}}}$

The answer must be rounded to one significant figure, the number of sig figs you have for the volume of water.