The correct option is A 0.61 M
Let the mass of the solution be 100 g.
So, 4.9 g of H3PO4 is present in 100 g of solution.
Moles of H3PO4
given massmolar mass=4.998=0.05 moles
Density=mass of solutionvolume of solution
⇒ volume of H3PO4=1001.22=81.97 mL
Molarity=moles of solute of H3PO4total volume of solution in mL×1000
Molarity = 0.0581.97×1000=0.61 M