Question

At a pressure of 760 torr and temperature of 273.15K, the indicated volume of which system is not consistent with the observation?

• A. 14 g of $N_2$ + 16 g of $O_2$; Volume = 22.4 L
• B. 4 g of He + 44 g of $C{O}_2$; Volume = 44.8 L
• C. 7 g of $N_2$ + 36 g of $O_3$; Volume = 22.4 L
• D. 17 g of $N{H}_3$ + 36.5 g of $HCl$, Volume = 44.8 L

Answer 1

The correct option is D 17 g of $N{H}_3$ + 36.5 g of $HCl$, Volume = 44.8 L

Given,

Pressure=1atm

temperature=$73.15K$

Now,For volume calculation,we know

1 mole of any gaseous mixture=$22.4L$

n mole of any gaseous mixture$=n\times 22.4L$

Volume of mixture$=n\times 22.4$

Now,

a)14g of $N_2$+16g of $O_2$

$=\frac{14}{28}+\frac{16}{32}=\frac{1}{2}+\frac{1}{2}=1$ mole

Volume = $22.4L$

b) 4g of $He+44g$ of ${CO}_2$

$=\frac{4}{4}+\frac{44}{44}=1+1=2$mole

Volume$=2\times 22.4=44.8L$

c)7g of $N_2$+36g of $O_3$

$=\frac{7}{28}+\frac{36}{48}=\frac{1}{4}+\frac{3}{4}=1$mole

Volume$=22.4L$

d)17g of ${NH}_3$+36.5g of HCl

$=\frac{17}{17}+\frac{36.5}{36.5}=2$mole

Volume of mixture$=44.8L$

Because ${NH}_3\left(g\right)+HCl\left(g\right)⟶{NH}_{4}Cl\left(s\right)$

Volume of solid<<volume of gas

Therefore, Volume of mixture is $<<44.8L$