The concentrated sulphuric acid that is peddled commercial is 95% $$\text{H}_2\text{SO}_4$$ by weight. If the density of this commercial acid is $$1.834\ \text{g cm}^{-3}$$, the molarity of this solution is

For molarity we always need “moles of solute present in 1 L of solution”.
Hence, first find the mass of the commercial acid that is contained in exactly 1 L of the solution and then convert that mass to moles.

Step 1: Mass of 1 L of the solution
Density $$\rho = 1.834\ \text{g cm}^{-3} = 1.834\ \text{g mL}^{-1}$$.
Volume considered $$V = 1\ \text{L} = 1000\ \text{mL}$$.
Mass of this volume $$m_\text{solution} = \rho V = 1.834 \times 1000 = 1834\ \text{g}$$.

Step 2: Mass of $$\text{H}_2\text{SO}_4$$ in that 1 L
The solution is 95 % by weight, i.e. $$95\ \text{g}$$ of $$\text{H}_2\text{SO}_4$$ per $$100\ \text{g}$$ of solution. Therefore,
$$m_{\text{solute}} = 0.95 \times m_\text{solution} = 0.95 \times 1834 = 1742.3\ \text{g}$$.

Step 3: Moles of $$\text{H}_2\text{SO}_4$$ in that 1 L
Molar mass $$M_{\text{H}_2\text{SO}_4} = 2(1) + 32 + 4(16) = 98\ \text{g mol}^{-1}$$.
$$n = \frac{m_{\text{solute}}}{M_{\text{H}_2\text{SO}_4}} = \frac{1742.3}{98} \approx 17.78\ \text{mol}$$.

Step 4: Molarity
Because the moles calculated are already for 1 L of solution,
$$\text{Molarity} = 17.78\ \text{mol L}^{-1} \approx 17.8\ \text{M}$$.

Therefore, the molarity of the commercial concentrated sulphuric acid is 17.8 M.

Option A which is: 17.8 M