Question

1.7 gm of PH₃ was kept in a sealed tube. Find out-
- the no of moles of PH₃
​-no of atoms present
-no of electrons present in it
-no of neutrons present in it
-no of molecules of PH₃

Answer 1

Number of moles of PH₃
$$\begin{gathered} Numberofmoles=\frac{Givenmass}{Molecularmass} \\ =\frac{1.7}{34}=0.05moles \\ Noofatoms=moles\times avogadronumber\times noofatomsinP{H}_3 \\ =0.05\times 6.023\times {10}^{23}\times 4 \\ =1.2046\times {10}^{23}atoms \\ Noofmolecules=moles\times avogadronumber \\ =0.05\times 6.023\times {10}^{23} \\ =0.3012\times 10{}^{23} \\ Noofelectrons=Numberofatoms\times numberofelectronsinP{H}_3 \\ =1.2046\times {10}^{23}\times (15+3)(noofelectronsinP=15andH=1) \\ =18.442\times {10}^{23}electrons \\ Noofneutrons=Numberofatoms\times numberofneutronsinP{H}_3 \\ =1.2046\times {10}^{23}\times (16+0)(noofneutronsinP=16andH=0) \\ =19.27\times {10}^{23}neutrons \end{gathered}$$