Question

Calculate the molality of $1L$ solution of $80\% $
${H_2}S{O_4}(W/V)$, Given that density of solution is $1.80gm{l^{ - 1}}$
A.$8.16$
B.$8.6$
C.$1.02$
D.$10.8$

Answer 1

Hint: In chemistry, the amount of substance calculated using the molar concept. Mole is the basic unit of this concept. It can calculate in solid as well as an aqueous solution too. Using Avogadro number large calculations is possible.

Complete step by step answer:
Before calculating the molality of the solution, let's have a look at the solution.
Solution - It is a mixture of two or more substances. In this mixture, solute and solvent are the two main parts. The solute is the substance that dissolves in another substance that is known as a solvent.
Molality - It defines how many moles of solute available in one kilogram of solvent. Its SI unit is moles per kilogram of solvent and the symbol is $m$. Using it we can identify the concentration of any solution. It comes from the concept of molarity or molar concentration.
Here, we have given,
volume - $1L$
Density - $1.80m{l^{ - 1}}$
${H_2}S{O_4}$ - $80\% $ which means $80g$ ${H_2}S{O_{}}$ in $100ml$ solution.
Therefore, $800g$ ${H_2}S{O_4}$ is present in $1000ml$ solution.
We know that Density = $\dfrac{{mass}}{{volume}}$
= $1.8$ =$\dfrac{{mass}}{{1000ml}}$
mass of solution = $1800g$
Molality = $\dfrac{{Moles\,of\,solute}}{{mass\,of\,solvent}}$
moles of ${H_2}S{O_4}$ = $2 + 32 + 64$ =$98$
molar mass of ${H_2}S{O_4}$ = $\dfrac{{mass\,}}{{moles}}$
= $\dfrac{{800}}{{98}}$ = $8.16$
mass of solvent = total mass $ - $ mass of solute
= $1800 - 800$
=$1L$
now, Molality = $\dfrac{{8.16}}{1}$
= $8.16molek{g^{ - 1}}$
Therefore, option $A$ is the correct answer.

Note:
Molarity - It calculates the concentration of a substance per unit volume of solution or we can say that how many moles per liter. Its symbol is $M$ and the SI unit is moles/liter. Avogadro Number- It is a constant number given by Amedeo Avogadro, its value is $6.022 \times {10^{23}}mo{l^{ - 1}}$.