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Question: One mole of element X has 0.444 times the mass of one mole of element Y. One atom element X has 2.96...

One mole of element X has 0.444 times the mass of one mole of element Y. One atom element X has 2.96 times the mass of one atom of 12C^{12}C. What is the atomic mass of Y?
A.80
B.15.77
C.46.67
D.40.0

Explanation

Solution

We can calculate the atomic mass of Y using the atomic mass of X and number of times of one mole of element X. We can calculate the atomic weight of X by multiplying one atom of X times with atomic weight of carbon. When we divide the atomic mass of X and the number of times of one mole of X we will get the atomic mass of Y.

Complete answer:
Given data contains,
One mole of element X has 0.444 times the mass of one mole of element Y.
One atom element X has 2.96 times the mass of one atom of 12C^{12}C.
We know that molecular mass of X is the product of 0.444 times the molecular mass of Y.
We can calculate the atomic mass of X as,
Atomic mass of X = 2.96×Atomic weight of carbon2.96 \times {\text{Atomic weight of carbon}}
We know that the atomic mass of carbon is twelve.
Atomic mass of X = 2.96×122.96 \times 12
Atomic mass of X = 35.5235.52
So, the atomic mass of X is 35.5235.52g.
So, 35.42=0.444×atomic weight of Y35.42 = 0.444 \times {\text{atomic weight of Y}}
Atomic weight of Y = Atomic weight of X0.444\dfrac{{{\text{Atomic weight of X}}}}{{0.444}}
Now we can substitute the value we get,
Atomic weight of Y = 35.420.444\dfrac{{35.42}}{{0.444}}
On simplifying we get,
Atomic weight of Y = 8080
So, we have calculated the atomic weight of Y is 8080.

So, the correct answer is Option A.

Note: Generally, the mass of an atom is known as atomic mass and SI of atomic mass is kilogram and it can also be expressed as Dalton. We can define atomicity as the total number of atoms that is found in a molecule. We can calculate atomicity using the molecular mass and atomic mass of the molecule. We can write the formula as,
Atomicity=Molecular massAtomic mass{\text{Atomicity}} = \dfrac{{{\text{Molecular mass}}}}{{{\text{Atomic mass}}}}.