Question

The Molarity of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ is $18{\text{ M}}$. Its density is $1.8{\text{ g m}}{{\text{L}}^{ - 1}}$. Hence molality is:
A.23 m
B.200 m
C.500 m
D.18 m

Answer 1

In molality mass of solvent is considered instead of volume of solution. Using density we will get the mass of solution. Using the definition of molarity, calculate the mass of solute. Subtract the mass of solution with mass of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ to get the mass of solvent.
Formula used: ${\text{Molality}} = \dfrac{{{\text{no}}{\text{. of moles }}}}{{{\text{mass of solvent}}({\text{g}})}} \times 1000$
We have multiplied with 1000 to convert mass into kilograms.
${\text{no}}{\text{. of moles}} = \dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$
Complete step by step answer:
Molality is defined as the number of moles of solute per unit mass of solvent taken in kilogram.
Molarity is defined as the moles of solute dissolved in 1 L of solution. We have been given the molarity of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ that is $18{\text{ M}}$. This means 18 moles of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ are present in the 1000 mL of solution.
${\text{moles of }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4} = \dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$
Mass of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ is 29 g. Molar mass of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ is $2 \times 1 + 32 + 4 \times 16 = 98$.
$\Rightarrow 18 = \dfrac{{{\text{mass}}}}{{{\text{98 g mo}}{{\text{l}}^{ - 1}}}}$
${\text{mass of }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4} = 18 \times 98 = 1764{\text{ g}}$
Volume of water is 1000 mL and density is given to us in question and hence we will calculate the mass of solution.
${\text{density}} = \dfrac{{{\text{mass}}}}{{{\text{volume}}}}$
$1.8{\text{ g m}}{{\text{L}}^{ - 1}} = \dfrac{{{\text{mass of solution}}}}{{{\text{1000 mL}}}}$
${\text{Mass of solution}} = 1800{\text{g}}$
We need mass of solvent that is water which can be calculated as:
Mass of solvent = mass of solution – mass of sulphuric acid
${\text{Mass of solvent}} = (1800 - 1764){\text{g = 36 g}}$
We will calculate the molality as following:
${\text{Molality}} = \dfrac{{{\text{18 }}}}{{36}} \times 1000$
${\text{Molality}} = 500{\text{ mol K}}{{\text{g}}^{ - 1}}$

Hence, the correct option is C.

Note:
Since the molarity involves volume so it is temperature dependent because volume changes with change in temperature. Since molality involves masses which are fixed and are not affected by temperature. Number of moles also consists of only masses and hence moles also remain unaffected with temperature and do is mole fraction. There is another concentration term known as normality, it also deals with volume and hence changes with temperature.