Question

if 280cm3 of hydrogen diffuses in 40 seconds, how long will it take for490cm3 of gas, x , whose vapors density is 25, to diffuse under thesame condition( relative molecular mass of hydrogen=2)

Answer 1

Explanation:

Hey user;

Here's answer to this question!;

Firstly we have to change units of Volume;

1 cm³ = 0.001 litres

280 cm³ = 0.28 litres

490 cm³ = 0.49 litres

Where;

$V_1 = Volume \: of \: hydrogen \\ \\ t_1 = time \: taken \: to \: diffuse \: by \: hydrogen. \\ \\ V_2 = Volume \: of \: gas \: ' x ' \\ \\ t_2 = time \: taken \: to \: diffuse \: by \: gas \: 'x'$

We should know the following formula to solve the following question;

$\boxed{\boxed{ \frac{r_1}{r_2} = \frac{\frac{V_1}{t_1}}{\frac{V_2}{t_2}} = \sqrt{ \frac{ d_2}{d_1}}} }$

Also Molecular weight = 2 × Vapour density

Molecule weight of gas 'x' is 50

$\frac{r_1}{r_2} = \frac{\frac{V_1}{t_1}}{\frac{V_2}{t_2}} = \sqrt{ \frac{ d_2}{d_1}} \\ \\ \frac{r_1}{r_2} = \frac{\frac{0.28}{40}}{\frac{0.49}{t_2}} = \sqrt{ \frac{ 50 }{2 }} \\ \\ \frac{7}{\frac{0.49}{t_2}} = \sqrt{ 25 } \\ \\ 7 \times t_2 = 2.45 \\ \\ \huge{\boxed{ \leadsto t_2 = 0.35 seconds}}$

Hope it helps you;

any doubt feel free to contact me!

Answer 2

Answer:

350 seconds

Explanation:

280cm^3 of hydrogen diffused in 40 seconds

490cm^3 of hydrogen will diffuse in x

490/280*40=70seconds

2/2*VD= Relative molecular mass of H2

; V.D of H2 gas =2/2=1

t1/t2=√d1/d2

70/t2=√1/25

70/t2=1/5

t2=70*5

=350 seconds

:490cm^3 of X would diffuse in 490/1.4=350 seconds.

I hope you find it useful.