KOH + HBr ---> KBr + H2O
0,3 moles of HBr ---in-------1000ml
x moles of HBr-------in------75ml
x = 0,0225 moles of HBr
according to the reaction: 1 mole of KOH = 1 mole of HBr
so
0,0225 moles of HBr = 0,0225 moles of KOH
0,0225 mole of KOH------in-----45ml
x moles of KOH -----------in------1000ml
x = 0,5 moles of KOH
answer: 0,5 mol/dm³ KOH (molarity)
Answer:

Explanation:
We know, 
where, R = 0.0821 L.atm/(mol.K), T is temperature in kelvin and
is difference in sum of stoichiometric coefficient of products and reactants
Here
and T = 311 K
So, ![K_{p}=(0.0111)\times [(0.0821L.atm.mol^{-1}.K^{-1})\times 311K]^{-1}=4.35\times 10^{-4}](https://tex.z-dn.net/?f=K_%7Bp%7D%3D%280.0111%29%5Ctimes%20%5B%280.0821L.atm.mol%5E%7B-1%7D.K%5E%7B-1%7D%29%5Ctimes%20311K%5D%5E%7B-1%7D%3D4.35%5Ctimes%2010%5E%7B-4%7D)
Hence value of equilibrium constant in terms of partial pressure
is 
Assume there is 100g of the substance at first