Question

How many moles and molecules are present in 10.0 g each of —

(A) H₂

(B) F₂

(C) HF

(D) F₂O

Answer 1

(A) Mol. wt. of $H_{2}=2 \cdot 0 g /$ mole

and mole $=\frac{\text { mass }}{\text { mol. wt. }}=\frac{10}{2}$ moles $=5.0$ moles.

Since, $1 mole =6.02 \times 10^{23}$ molecules

$\therefore 5$ moles $=5 \times 6.02 \times 10^{23}$ molecules $=3.01 \times 10^{24}$ molecules.

(B) Mol. wt. of $F_{2}=19 \times 2=38 g / mole$

$\therefore \quad$ No. of mole of $F_{2}=\frac{\text { mass }}{\text { mol. wt. }}=\frac{10}{38}$ mole $=0.263$ mole.

and no. of molecules $=0.263 \times 6.02 \times 10^{23}$ molccules $= 1 \cdot 58 \times 10^{23}$ molecules.

(C) Mol. wt. of $H F=1+19=20 g / mole$

Therefore, 1.0. of molc of $H F$ in $10 g =\frac{10}{20}$ molc $=0.5$ mole.

and also no. of molecules $=0.5 \times 6.02 \times 10^{23}$ molecules $= 3 \cdot 0 1 \times 10^{23}$ molecules.

(D) Mol. wt. of $F _{2} O =19 \times 2+16=54 g /$ mole

No. of mole in $10 g$ of $F_{2} O=\frac{\text { mass }}{\text { mol. wt. }}=\frac{10}{54}$ mole $= 0 \cdot 185 mole$.

and no. of molecules $=0.185 \times 6 \cdot 02 \times 10^{23}$ molecules $= 1 \cdot 1 1 \times 10^{23}$ molecules.