Decreasing order (first having highest and then others following it) of mass of pure NaOH in each of the aqueous solution:(i) 50 g of 40% (w/W) NaOH(ii) 50 mL of 50% (w/V) NaOH [dsoln. = 7.2 g / mL](iii) 50 g of 15 M NaOH [dsoln. : 1g,/mL]

(i) Mass of NaOH =50×40100=20g(ii) wvNaOH→50 g NaOH in 100mLSo, ln 50 mL mass of NaOH will be =25 g(iii)  50 g of 15 M NaOH(d =1g/mL)no.of moles n = MV = WGMW                   W =50×15×40       =300                               Order (iii) > (ii) > (i)