Answer:- 14.9 M
Solution:- Given commercial sample of ammonia is 28% by mass. Let's say we have 100 grams of the sample. Then mass of ammonia would be 28 grams.
Density of the solution is given as 0.90 grams per mL.
From the mass and density we could calculate the volume of the solution as:
= 111 mL
Let's convert the volume from mL to L as molarity is moles of solute per liter of solution.
= 0.111 L
Now, we convert grams of ammonia to moles on dividing the grams by molar mass. Molar mass of ammonia is 17 gram per mole.
= 1.65 mole
To calculate the molarity we divide the moles of ammonia by the liters of solution:
= 14.9 M
So, the molarity of the given commercial sample of ammonia is 14.9 M.
The density is 3.144 g / cm^3.
<u>Explanation</u>:
If effective number of atom in NaCl type structure, z = 4
a = 705.2 pm ⇒ In centimeter = 705.2 
10^-10
Na = 6.023 10^23
density = (molecular weight) (z) / (Na) (a^3)
where molecular weight of KI is 166 g,
Z represents the atomic number
density = (molecular weight) (z) / (Na) (a^3)
= (166 4) / (6.023 10^23) (705.2 10^-10)
density = 3.144 g / cm^3.
Answer:
1.51 X 10^23 ions
Explanation:
The number of ions in 17.1 gm of aluminum sulphate Al2 (SO4)3 =….. [Molar mass of Al2 (SO4)3 = 342 gm]
in one molecule of Al2(SO4)3 there are 5 ions 2 aluminum and 3 sulfate ions
in 2 molecules there are 2X5= 10 ions
in 10 molecules there are 10X5 = 50 ions
molar mass of Al2(SO4)3 = (2 X 26.98) +( 3 X 32.1) + (3 X 4 X 16.0 ) =342.gms = 17.1/342 =0.0500 moles
1 mole =6.02 X 10^23 molecules ( see Avogadros number)
0.0500 moles = 0.0500 X 6.02 X 10^23 molecules =
0.301 X 10^23 molecules = 3.01 X 10^22 molecules
We determined that each molecule of Al2(SO4)3 has 5 ions
so 3.01 X10^22 molecules have 5 X 3.01 X 10^22 ions =
15.05 X 10^22 ions = 1.51 X 10^23 ions
First and foremost, they are completely different substances with each exhibiting unique properties. Both have different atoms involved on their structures which is the cause of the differing properties.
