Question

A sulphuric acid solution contains \[98\% \left( {m/v} \right){H_2}S{O_4}\]. The density of the solution is $1.98g/c{m^3}$. The molarity and molality respectively of the solution are:
A.$19.8M\& 500m$
B.$10M\& 1.5m$
C.$10M\& 10m$
D.$1.0M\& 1m$

Answer 1

: To solve this question, we should know the basic concepts of chemistry. Molarity and molality are basic quantities in chemistry. Both molarity and molality are dependent on the number of moles. So we will aim to calculate the number of moles of solute first.
Formula Used:
Molarity
$M = \dfrac{n}{V}$
Where $M$ represents Molarity,
$n$ represents the number of moles of solute,
$V$ represents the volume of solution in ml.
Molality
$m = \dfrac{n}{W}$
Where $m$ represents molality,
$n$ represents the number of moles of solute,
$W$ represents the weight of the solvent in gram.

Complete step by step answer:
First, we will understand the meaning of a sulphuric acid solution containing \[98\% \left( {m/v} \right){H_2}S{O_4}\]. It simply means that the sulphuric acid solution contains $98g$ \[{H_2}S{O_4}\]. We are assuming here $100g$of the solution.
Now we can easily calculate the weight of the solvent ${W_{solvent}} = 100 - 98 = 2g$. The solvent is considered water.
Now we have given quantities as $d = 1.98g/c{m^3},{W_{{H_2}S{O_4}}} = 98g,{W_{solution}} = 100g,{W_{solvent}} = 2g$
We can easily calculate the volume of solution using the formula, $d = \dfrac{{Mass}}{{Volume}}$ substituting values in the formula we get,
$ \Rightarrow 1.98 = \dfrac{{100}}{{Volume}}$
$ \Rightarrow Volume = 50.5c{m^3}$
Hence, the volume of the solution is $Volume = 50.5c{m^3}$
Now we will calculate the number of moles of solute using the formula, $n = \dfrac{w}{{{M_{wt.}}}}$. We have $w = 98g,{M_{{H_2}S{O_4}}} = 2 \times 1 + 32 + 16 \times 4 = 98g/mol$ to substitute the values in the formula we get,
$ \Rightarrow n = \dfrac{{98}}{{98}} = 1$
Therefore, the number of moles of solute is $n = 1$.
Now we have all the quantities to calculate the molarity and molality. For molarity,
$M = \dfrac{n}{V},n = 1,Volume = 50.5c{m^3} = 50.5ml$
Substituting value in the formula we get,
$M = \dfrac{1}{{50.5}} \times 1000 = 19.8mol/L$
For molality, $m = \dfrac{n}{W},n = 1,{W_{Solvent}} = 2g$, substituting the values in the formula we get,
$m = \dfrac{1}{2} \times 1000 = 500mol/kg$
Therefore, the values of molarity and molality are $19.8M\& 500m$.

Therefore, the correct option is (A).

Note:
Molality is preferred over molarity as molarity is temperature-dependent and molality is temperature independent. Thus molarity changes with change in temperature as molarity is inversely proportional to volume.