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

$(a)\;\frac{y}{2} \\ (b)\;\frac{y}{4} \\(c)\;y \\(d)\;2y$

No. of moles of $A=\large\frac{x}{40}$
Number of atoms of $A= \large\frac{x}{40}$$\times NA =y No. of moles of B =\large\frac{2x}{80}=\frac{x}{40} No. of atoms of B =\large\frac{x}{40}$$NA$
Hence c is the correct answer.