Question

A 19.6 g of a given gaseous sample contains 2.8 g of molecules $d=0.75g{L}^{-1}$ 11.2 g of molecules $(d=3g{L}^{-1})$ and 5.6 g of molecules $(d=1.5g{L}^{-1})$. All density measurements are made at STP. Calculate the total number of molecules (N) present in the given sample. Report your answer in $N\times {10}^{23}$.
Assume Avogardro's number as $6\times {10}^{23}$.

• A. 3
• B. 2
• C. 4
• D. 2.5

Answer 1

The correct option is A 3

Total weight = 19.6 g
(a) $2.8g$ of molecules
$d=0.765g{L}^{-1}$ $\text{Volume}=\frac{\text{Mass}}{d}=\frac{2.8}{0.75}L$ at STP
1 mol $\equiv 22.4$ L at STP
Moles $=\frac{2.8}{0.75}\times \frac{1}{22.4}$ Molecules $=\frac{2.8}{0.75}\times \frac{1}{22.4}\times 6\times {10}^{23}=1\times {10}^{23}$
(b) $11.2g$ of molecules
$d=3g{L}^{-1}$
$\text{Volume}=\frac{\text{Mass}}{d}=\frac{11.2}{3}L$ at STP
1 mol $\equiv 22.4$ L at STP
Moles $=\frac{11.2}{3}\times \frac{1}{22.4}$ Molecules $=\frac{11.2}{3}\times \frac{1}{22.4}\times 6\times {10}^{23}=1\times {10}^{23}$

(c) $5.6g$ of molecules
$d=1.5g{L}^{-1}$
$\text{Volume}=\frac{\text{Mass}}{d}=\frac{5.6}{1.5}L$ at STP
1 mol $\equiv 22.4$ L at STP
Moles $=\frac{5.6}{1.5}\times \frac{1}{22.4}$ Molecules $=\frac{5.6}{1.5}\times \frac{1}{22.4}\times 6\times {10}^{23}=1\times {10}^{23}$
Molecules $=1\times {10}^{23}$
Total number of molecules $\left(a\right)+\left(b\right)+\left(c\right)=3\times {10}^{23}\Rightarrow N=3$