Given:
- Oxide of metal contains 30% oxygen by weight.
- Atomic ratio of metal and oxygen is 2:3.

To determine: Atomic weight of metal.

Solution:
Assume the following:
- Let the atomic weight of metal be 'x'.
- Let the atomic weight of oxygen be 'y'.

From the given atomic ratio of metal and oxygen, we can write:
2x : 3y

As per the law of definite proportions, the percentage composition of the oxide will always be constant. Hence, we can write:
% of metal in oxide = (atomic weight of metal / molecular weight of oxide) x 100
% of oxygen in oxide = (atomic weight of oxygen / molecular weight of oxide) x 100

Given that the oxide contains 30% oxygen by weight, we have:
% of oxygen in oxide = 30

Substituting the values in the above equation, we get:
(y / (x + y)) x 100 = 30
=> y = 0.3(x + y)

Also, we know that:
Atomic weight of metal + Atomic weight of oxygen = Molecular weight of oxide
=> x + y = molecular weight of oxide

Again, from the given atomic ratio, we can write:
2x : 3y = molecular weight of oxide / (2x + 3y)

Substituting the value of 'y' from equation (1), we get:
2x : 3(0.3x + 0.3y) = molecular weight of oxide / (2x + 3(0.3x + 0.3y))

Simplifying the above equation, we get:
2x : 0.9x + 0.9y = molecular weight of oxide / (2.9x + 0.9y)

Multiplying both sides by (2.9x + 0.9y), we get:
2x(2.9x + 0.9y) = molecular weight of oxide x 2
=> 5.8x^2 + 1.8xy = 2x

Substituting the value of 'y' from equation (1) and simplifying, we get:
5.8x^2 + 0.54x^2 = 2x
=> 6.34x^2 = 2x
=> x = 56

Therefore, the atomic weight of metal is 56, which is given in option (B).