3.15 grams of solute is present in 200ml of 0.25 M solution. The solute may be

$Molarity, M = \frac{no. of moles of solute}{vol. of solution (in L)}$

$\mathrm{no}. \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{solute}, \mathrm{n} = \frac{\mathrm{given} \mathrm{mass}}{\mathrm{molar} \mathrm{mass}}$

Molarity of a solution is defined as the amount of solute dissolved in 1L of solution.

M = 0.25 M, grams of solute = 3.15 g ,vol. of solution = 200 mL = 0.2 L

So,  $0.25 = \frac{{\displaystyle \raisebox{1ex}{$3.15$}\!\left/ \!\raisebox{-1ex}{$MM$}\right.}}{0.2}$ ; molecular mass = 63 g/mol

H₂C₂O₄ = 90 g/mol ;HCl = 36.5 g/mol ; H₂SO₄ = 98 g/mol ; HNO₃ = 63 g/mol

Hence, correct option is: (D) ${\mathrm{HNO}}_3$