The density (in g $$mL^{-1}$$) of a 3.60 M sulphuric acid solution that is 29% $$H_2SO_4$$ (Molar mass = 98 g $$mol^{-1}$$) by mass will be

Molarity is defined as moles of solute present in exactly $$1 \text{ L}$$ of solution.

For a $$3.60 \text{ M}$$ $$H_2SO_4$$ solution:
moles of $$H_2SO_4$$ in $$1 \text{ L} = 3.60$$

Molar mass of $$H_2SO_4 = 98 \text{ g mol}^{-1}$$.
Mass of acid present in $$1 \text{ L}$$ solution is therefore
$$3.60 \times 98 = 352.8 \text{ g}$$.

The solution is given to be $$29\%$$ (w/w), meaning
$$\frac{\text{mass of } H_2SO_4}{\text{mass of solution}} = 0.29$$.

Let the mass of the whole solution occupying $$1 \text{ L}$$ be $$m_{\text{soln}}$$ g. Then $$0.29 \, m_{\text{soln}} = 352.8 \text{ g}$$

$$m_{\text{soln}} = \frac{352.8}{0.29} \approx 1.216 \times 10^{3} \text{ g}$$.

Density, $$\rho = \frac{\text{mass of solution}}{\text{volume of solution}}$$

Volume considered = $$1 \text{ L} = 1000 \text{ mL}$$, hence
$$\rho = \frac{1.216 \times 10^{3} \text{ g}}{1000 \text{ mL}} = 1.22 \text{ g mL}^{-1}$$.

Option C which is: 1.22