1. Given data

• Molarity of solution $= 3.60\,\text{M}$
• Mass percent of $H_2SO_4 = 29\%$
• Molar mass of $H_2SO_4 = 98\,\text{g mol}^{-1}$

We need to find the density of the solution in $\text{g mL}^{-1}$.

2. Take 1 L of solution

Since molarity is given, let us consider exactly $1\,\text{L}$ of solution.

Then moles of $H_2SO_4$ in $1\,\text{L}$ are:

$n = 3.60\,\text{mol}$

So mass of $H_2SO_4$ in $1\,\text{L}$ solution is:

$m_{H_2SO_4} = 3.60 \times 98 = 352.8\,\text{g}$

3. Use mass percent relation

The solution is $29\%$ $H_2SO_4$ by mass, so

$\frac{\text{mass of solute}}{\text{mass of solution}} \times 100 = 29$

Thus,

$\frac{352.8}{\text{mass of solution}} \times 100 = 29$

So mass of $1\,\text{L}$ solution is:

$\text{mass of solution} = \frac{352.8 \times 100}{29}$

$= 1216.55\,\text{g}$

4. Calculate density

Volume of solution $= 1\,\text{L} = 1000\,\text{mL}$

Therefore density is:

$\rho = \frac{1216.55}{1000} = 1.21655\,\text{g mL}^{-1}$

Approximating,

$\rho \approx 1.22\,\text{g mL}^{-1}$

5. Check options

• A: $1.45$ ❌
• B: $1.64$ ❌
• C: $1.88$ ❌
• D: $1.22$ ✅

So the correct option is D.

6. Comparison with stored correct answer

Stored correct answer: D

My derived answer: D

They agree.