Question

Two samples of sulphuric acid are labelled as $1M$ $H_{2}S{O}_4$ and $1m$ $H_{2}S{O}_4$. The density of solution is $1.15g/cc$, then which among the following is more concentrated?

• A. $1M{H}_{2}S{O}_4$
• B. $1m{H}_{2}S{O}_4$
• C. Both are equal
• D. Cannot be predicted

Answer 1

The correct option is B $1m{H}_{2}S{O}_4$

Given that

Molarity of the $1^{st}$ sample, $M_1=$ 1 M

Molality of the $2^{nd}$ sample, $m_2=$ 1 m

Density of the solution, d $=$ 1.15 g/cc

Assuming the volume of $1^{st}$ sample taken, $V_1=$ 1 lit $=$ 1000 cc

Molar mass of the solute $H_{2}S{O}_4$, $M^{\ast }=$ 98 g/mol

Now,

Mass of the $1^{st}$ sample solution, $W_1=$ Density $\times$ Volume $=$ d $\times V_1$

$=$ 1.15 g/cc $\times$ 1000 cc$=$ 1150 g.

Number of moles of solute in $1^{st}$ sample, $n_1=$ Molarity $\times$ Volume $=M_1\times V_1$

$=$ 1M $\times$ 1 lit $=$ 1 mol.

Mass of the solute of the $1^{st}$ sample, $w_2=$ No. of moles $\times$ Molar mass of the solute

$=n_1\times M^{\ast }=$ 1 mol $\times$ 98 g/mol $=$ 98 g.

Mass of the solvent of the $1^{st}$ sample, $w_1=$ Mass of the solution $-$ Mass of the solute

$=$ 1150 g $-$ 98 g $=$ 1052 g $=$ 1052/1000 kg $=$ 1.052 kg.

$\therefore$ Molality of the $1^{st}$ sample, $m_1=\frac{molesofsolute}{massofthesolventinkg}=\frac{n_1}{w_1}=\frac{1mol}{1.052kg}\approx$ 0.951 m

Since, $m_1<m_2$ i.e. 0.951 m $<$ 1 m

So, 1 m $H_{2}S{O}_4$ is more concentrated than 1 M $H_{2}S{O}_4$

$\therefore$ (B) option is correct.