What is molaritiy of the soiution prepared by dissolving 75.5 g of pure KHO in 540 ml of solution?

Molar mass of KOH = (39.0 + 16.0 + 1.0) g/mol = 56.0 g/mol
No. of moles of KOH = (75.5 g) / (56.0 g/mol) = 1.35 mol

Molarity of the solution = (1.35 mol) / (540/1000 L) = 2.50 M

First of all, what is 'KHO' ? If you mean potassium hydroxide , then the formula is 'KOH'!!!!

All hydroxides/alkalis are written in this form; it is internationally recognised.

First calculate the 'Mr' of KOH
K x 1 = 39 x 1 =39
O x 1 = 16 x 1 = 16
H x 1 = 1 x 1 = 1
39 + 16 + 1 = 56

Hence moles (KOH) = 75.5 / 56 = 1.3482 moles.
This is dissolved in 540 mL = 0.54 dm^3
Hence
Molarity is 1.3482 / 0.54 = 2.4966... mol dm^-3 = ~ 2.50 mol dm^-3 = 2.50 M

(75.5 g KOH) / (56.10564 g KOH/mol) / (0.540 L) = 2.49 mol/L KOH