Question

The atomic weights of two elements $A$ and $B$ are $40$ and $80$ respectively. If $xg$ of A contains $y$ atoms, how many atoms present in $2xg$ of B?

• A. $\frac{y}{2}$
• B. $y$
• C. $2y$
• D. $\frac{y}{4}$

Answer 1

The correct option is B $y$

$\begin{array}{ccc}& A& B\\ \text{Atomic mass}& 40& 80\\ \text{Given mass (g)}& xg& 2xg\\ \text{No.of moles}mol& \frac{x}{40}& \frac{2x}{80}\\ \text{No.of atoms}& \frac{x}{40}{N}_A& \frac{2x}{80}{N}_A\end{array}$

$No.ofmoles=\left(\frac{mass}{molarmass}\right)$

No. of atom = no. of moles $\times N_A$

And given no.of atoms in $xg$ of $A=y$

$\therefore \frac{x}{40}\times N_A=y$

$x=\frac{40y}{N_A}$

And no.of atoms in $2xg$ of $B=\frac{2x}{80}{N}_A$

Substituting the value of $x$ in the above equation:

$\frac{2\times 40y\times N_A}{80\times N_A}$

$=y$

Hence, $y$ atoms are present in $2xg$ of B.

Option (c) is the correct answer.