Question

If $(N,\text{ }P\text{ },\text{ }As)$ was Dobereiner’s triads then what would be the absolute error in atomic mass of phosphorus as compared to the original mass?
Atomic mass: Nitrogen =$14\text{ }amu$, Arsenic =$74.9\text{ }amu$, Phosphorous =$31\text{ }amu$
A. $13.45$
B. $12.45$
C. $15.55$
D. $16.55$

Answer 1

Hint : To solve this problem we first know the atomic mass of the elements given in this question. Here we apply this formula $M(B)=\dfrac{\left( M\left( A \right)+M\left( C \right) \right)}{2}$ , where B is the second element, A is the first element, C is the third element.

Complete step-by-step solution:
According to scientist Dobereiner, the arithmetic mean of the atomic masses of the first and third element in a triad would be nearly adequate for the atomic mass of the second element within the triad.
We assume that $N,P$, $As$ follow Dobereiner’s triads, therefore,

& M(P)=\dfrac{\left( M\left( N \right)+M\left( As \right) \right)}{2} \\\ & \Rightarrow M(P)=\dfrac{\left( 14+74.9 \right)}{2} \\\ & \Rightarrow M(P)=44.45 \\\ \end{aligned}$$ $$\Rightarrow 44.45-31=13.45$$ **Therefore correct option is A. $$13.45$$** **Note:** There are some limitations present in Dobereiner’s triads i.e. recently discovered elements couldn’t fit into triads and only five triads are found. The first triad was found by Scientist Dobereiner and was composed of alkaline earth metals calcium, strontium, and barium.