Question

Find the number of moles of phosphorus in $92.9$ g of phosphorus assuming that molecular formula of phosphorus in $P_4$. Also, determine the number of atoms and molecules of phosphorus in the sample.

• A. $3$ mol, $7$ $\times {10}^{23}$ molecules of $P_4$, $18.04$ $\times {10}^{23}$ atoms of $P$
• B. $0.75$ mol, $4.52$ $\times {10}^{23}$ molecules $P_4$, $12$ $\times {10}^{23}$ atoms $P$
• C. $0.75$ mol, $4.52$ $\times {10}^{23}$ molecules $P_4$, $18.04$ $\times {10}^{23}$ atoms $P$
• D. none of these

Answer 1

Answer: (C)

$0.75$ mol, $4.52$ $\times {10}^{23}$ molecules $P_4$, $18.04$ $\times {10}^{23}$ atoms $P$

Atomic mass of $P$ is $31$g. So, number of moles of $P$ in $93$ g = $\frac{93}{31\times 4}$ = $0.75$ moles.
Number of molecules of $P_4=$ $4.52$ $\times {10}^{23}$
Also, number of atoms = $4.52$$\times 4\times {10}^{23}$ = $18.04$ $\times {10}^{23}$