The correct option is B 18 M
Taking the basis as 100 g of the solution, mass of the acid is 98 g since % w/w=98%.
The density of solution is 1.8 g ml−1
Molar mass of the acid, Mo = 98 gm mol−1
Molarity:
=molesVsolution in L=%w/w(g)Mo(g/mol)×d(g/mL)100(g)×1000(mL)1(L)=% w/wMo×d×10
M=9898×1.8×10=18 M