Answer : The molar mass of the unknown gas will be 79.7 g/mol
Explanation : To solve this question we can use graham's law;
Now we can use nitrogen as the gas number 2, which travels faster than gas 1;
So, 167 / 99 = 1.687 So the nitrogen gas is 1.687 times faster that the unknown gas 1
We can compare the rates of both the gases;
So here, Rate of gas 2 / Rate of gas 1 = 
Now, 1.687 = square root [
]
When we square both the sides we get;
2.845 = (molar mass 1) / (28.01 g/mol N2)
On rearranging, we get,
2.845 X (28.01 g/mol N2) = Molar mass 1
So the molar mass of unknown gas will be = 79.7 g/mol
The way you want to find the percent composition would be by breaking down the problem like so:
K= atomic mass of K which is 39.098
Mn = atomic mass of Mn which is 54.938
O= atomic mass of o which is 15.999
Then you want to add 39.098+ 54.938+ 15.999 and you get 110.035 which is the molar mass for KMnO
Then you want to take each molar mass and then divide it 110.035 and multiply by 100
Ex. K = 39.098/ 110.035 and the multiply what you get by a 100
You do this for the other elements as well good luck!
Reaction of dissociation: Ag₂SO₄ → 2Ag⁺ + SO₄²⁻.
m(Ag₂SO₄) = 4 g.
V(Ag₂SO₄) = 1 l.
n(Ag₂SO₄) = m(Ag₂SO₄) ÷ M(Ag₂SO₄).
n(Ag₂SO₄) = 4 g ÷ 311,8 g/mol.
n(Ag₂SO₄) = 0,0128 mol.
n(Ag⁺) = 2 · 0,0128 mol = 0,0256 mol.
n(Ag₂SO₄) = n(SO₄²⁻) = 0,0128 mol.
c(Ag⁺) = n ÷ V = 0,0256 mol ÷ 1 l = 0,0256 mol/l.
Ksp = c(Ag⁺)² · c(SO₄²⁻).
Ksp = (0,0256 mol/l)² · 0,0128 mol/l.
Ksp = 8,3·10⁻⁶.
