Question

A compound which contains one atom of $X$ and two atoms of $Y$ for each three atoms of $Z$ is made by mixing $5.0g$ of $X$, $1.15\times {10}^{23}$ atoms of $Y$ and $0.03g-atoms$ of $Z$. If only $4.40g$ of the compound results, then the value of atomic mass of $Y$ is (The atomic masses of $X$and $Z$ are $60$ and $80$, respectively and take avogadro number as $6\times {10}^{23}$.)

• A. 70

Answer 1

The correct option is A 70
According to question: molecule is $X{Y}_2{Z}_3$
Let atomic mass of $Y=A_y$
Moles of $X$-atoms = $\frac{givenmass}{molarmass}$ = $\frac{5.0}{60}=0.083$ moles
Moles of $Y$-atoms = $\frac{1.15\times {10}^{23}}{6\times {10}^{23}}$
= $0.192$ moles
Moles of $Z$-atoms = $0.03$

Hence, Limiting reagent is $Z$ and with respect to $Z$, $0.01$ mole of $X{Y}_2{Z}_3$ will be formed.
Using formula:
Mole = $\frac{givenmass}{molarmassofcompound}$
⇒ $0.01=\frac{4.40}{60}+2A_y+3\times 80$
⇒ $A_y=70$