Question

Molarity of ${{H}_{2}}S{{O}_{4}}$ is $0.8$ and its density is $1.06g/c{{m}^{3}}$. What will be its concentration in terms of molality?
(A) $0.54m$
(B) $0.84m$
(C) $0.814m$
(D) $0.514m$

Answer 1

Hint: Molality is related to solvent rather than to the solution. It is a ratio where the denominator is in kilograms.

Complete step-by-step answer:
Let’s start from the basics and understand each of the terms given above. We first look into molarity.
Molarity is defined as the number of moles of solute present in one litre of solution. Molality is defined as the number of moles of solute present in one kilograms of the solvent. And last but not the least density is the mass per unit volume of any given substance.
It is given that molarity of ${{H}_{2}}S{{O}_{4}}$ is $0.8$, which means $0.8$ moles of ${{H}_{2}}S{{O}_{4}}$ is present in one litre of the solution. As one litre is equal to a thousand millilitres we can say that $0.8$ moles of ${{H}_{2}}S{{O}_{4}}$ is present in one thousand millilitres of the solution. The equations are as below:

& 1l\to 0.8\text{ moles} \\\ & \Rightarrow \text{1000ml}\to \text{0}\text{.8 moles} \\\ \end{aligned}$$ The density of the solution is given as $1.06g/c{{m}^{3}}$or $1.06g/ml$. Mass of the solution is: $$\begin{aligned} & Density(D)=\dfrac{Mass(M)}{Volume(V)} \\\ & \Rightarrow D=\dfrac{M}{V}\Rightarrow M=D\times V\Rightarrow M=1.06g/ml\times 1000ml \\\ & \Rightarrow M=1060g \\\ \end{aligned}$$ The above calculations were simply based upon the formula of density. The mass of the solution is therefore $1060g$. To find out molality we require the mass of solvent. If we subtract the mass of solute (which in this case is${{H}_{2}}S{{O}_{4}}$) from the total mass of the solution we can get the mass of solvent. $$\begin{aligned} & Mass\text{ of solute = number of moles of solute}\times \text{molar mass of solute} \\\ & \Rightarrow \text{Mass of }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}=0.8\times 98=78.4g \\\ \end{aligned}$$ Therefore mass of solvent is: $$\begin{aligned} & Mass\text{ of solution - Mass of solute} \\\ & =1060-78.4 \\\ & =981.6g \\\ \end{aligned}$$ Now molality is moles of solute present in thousand grams of solvent. By unitary method the following is solved: $$Molality=\dfrac{0.8}{981.6}\times 1000=0.814$$ So the molality of ${{H}_{2}}S{{O}_{4}}$ is $0.814m$ which makes option (C) the correct answer. Note: Even if you do not remember the formula of molality it can still be calculated if you have your basics clear, as done above. Caution should be taken to not take the density given in the question to be that of the solvent. In most of the cases it is the solution from which we have to calculate the mass of the solvent.