Question

The reaction between gaseous NH3 and HBr produces a white solid NH4Br . Suppose the NH3 and HBr are introduced simultaneously in to the opposite ends of an open tube of 1 metre long. Where would you expect the white solid to form?​

Answer 1

$\huge❥︎\huge \large\mathbb{\pink{\underline{\underline{Answer}}}}$

Let, the distance of white solid from $NH_3$ end be 'x'.

Let, the distance of white solid from HBr end be (100-x).

The rates of diffusion should be in proportion to distances.

$\huge\frac{r_1}{r_2}$ = $\huge\frac{x}{(100-}$ = $\huge\sqrt\frac{MHBr}{MNH3}$

Molecular mass of HBr = 1+80 = 81

Molecular mass of NH3 = 14+3 = 17

So, $\huge\frac{x}{(100-x)}$ = $\huge\sqrt\frac{81}{17}$

$\rightarrow$ $\huge\frac{x}{(100-x)}$

So, x = 100×2.18-2.18x

$\Rightarrow$ $x$+2.18$x$= 100×2.18

$\Rightarrow$ 3.18$x$ = 100×2.18

$\Rightarrow$ $x$= $\huge\frac{100×2.18}{3.18}$

$x$ = 68.55cm

$\therefore$ The solid forms 68.55cm from $NH_3$.

$\sf\blue{hope\:it\:helps}$

Answer 2

$\bigstar\huge\bf\underline{\underline{Answer:-}}$

$\bf Let \ distance \ of \ white \ solid \ from \ NH_{3} \ end = \ x \ cm$

$\bf The \ distance \ of \ white \ solid \ from \ HBr \ end \ = \ (100-x)cm$

$\bf Rates \ of \ diffusion \ shall \ be \ proportional \ to \ these \ distances$

$\bf\implies \frac{r_{1}}{r_{2}}=\frac{x}{(100-x)}=\sqrt{\frac{M_{HBr}}{M_{NH_{3}}}}$

$\bf Molecular \ mass \ of \ HBr=1+80$

$\bf\implies 81$

$\bf Molecular \ mass \ of \ NH_{3}=14+3$

$\bf\implies 17$

$\bf \frac{x}{(100-x)}=\sqrt{\frac{81}{17}}\implies2.18$

$\bf Here,$

$\bf\implies x=100\times2.18-2.18x$

$\bf\implies x=\frac{100\times2.18}{3.18}$

$\bf\boxed{\boxed{\bf x=68.55cm}}$