What is the mass of KOH in 150cm³ of semi-molar aqueous solution?

Potassium hydroxide, KOH, is a strong base; and thus dissociates completely in aqueous medium as:

KOH(aq) → K+(aq) + OH-(aq).

Therefore [KOH]=[OH-]=0.0020M =2.0x10^(-3) M. ——————(1)

In an aqueous medium the equilibrium constant for H2O is

Kw=[H3O+][OH-] =1x10^(-14)M^2.—(2).

(a) From equation (2) we can find [H3O+] by:

[H3O+]=1x10^(-14)M^2/[OH-],

and substituting value for [OH-], we have:

[H3O+]={1x10^(-14)/0.0020} M

[H3O+]=500x10^(-14)M=5.00x10^(-12)M

Thus the required concentration of H3O+ is 5.00x10^(-12)M

(b) To find the pH use the definition

pH=-log[H3O+] and thus,

pH=-log[5.00x10^(-12)]=12-log5,

pH=12–0.6990=11.3.

Thus the required pH of the solution is 11.3.

(c) The required OH- concentration is as earlier stated in equation (1) as 2.0x10^(-3)M.

Note: Units of Kw are mol^2/l^2 or M^2.

I am assuming that the OP has taken the density of water as 1.2g/ml

Molality =mol of solute/ kg of solvent

First I will find the molar mass of solute which is the NH3

N= 14 g

H x 3 = 1 x 3 = 3g

Molar mass of NH3 = 17g

Moles of solute = mass of solute/molar mass of solute

Moles of NH3 =0.85/17= 0.05mol of NH3

Now I will calculate the mass of solvent in kg

I am given the volume of water which is 100 ml and I am given the density of water which is 1.2g/ml. Since

Mass = volume x density

Mass of water will be = 100 x 1.20= 120g

I must convert the 120g into kg.

I know that 1000g=1kg, I only have 120g so I know that I will have less than 1kg.

1000g = 1kg

120g = xkg

cross multiplying gives

1000x =120

Divide both sides by 1000

1000x/1000 = 120/1000

x = 0.120kg

Back to the definition of molality

Molality = moles of solute/ kg of solvent

Molality of NH3 = 0.05/0.120 = 0.417molal