Question

Same volume of two gases X and Y diffuses through same pin holes in 15 and 25 seconds respectively under identical conditions. If the molar mass of X is 30 g/mole
then the molar mass of Y will be:-
(a) 25.5 g/mol
(b) 83.3 g/mol
(c) 55 g/mol
(d) 64 g/mol​

Answer 1

AnswEr :-

• b) 83.3 g/mol

GivEn :-

• Same volume of two gases X and Y diffuses through same pin holes in 15 and 25 seconds respectively under identical conditions.
• Molar mass of X is 30 g/mol

To Find :-

• Molar mass of Y

Solution :-

We will use the following formula here to solve this one :-

$\implies \boxed{ \sf \: \dfrac{T_{x} }{T_{y}} = \sqrt{ \dfrac{M_{x}}{M_{y}} } }$

where,

• Tx = Time taken by X gas to be diffused through the hole
• Ty = Time taken by Y gas to be diffused through the hole
• Mx = Molar mass of X gas
• My = Molar mass of Y gas

➦ Now, substituting the values,

$\implies \boxed{ \sf \: \dfrac{15 }{25} = \sqrt{ \dfrac{30}{M_{y}} } }$

$\implies \boxed{ \sf \: \dfrac{3 }{5} = \sqrt{ \dfrac{30}{M_{y}} } }$

➦ Squaring both sides,

$\implies \boxed{ \sf \: \dfrac{9 }{25} = { \dfrac{30}{M_{y}} } }$

$\implies \sf \: M_{y} = \dfrac{25 \times 30}{9}$

$\implies \sf \boxed{ \red{ \bold{ M_{y} = 83.33}}}$

Answer 2

Answer:

$AnswEr :-</p><p></p><p>b) 83.3 g/mol</p><p></p><p>GivEn :-</p><p></p><p>Same volume of two gases X and Y diffuses through same pin holes in 15 and 25 seconds respectively under identical conditions.Molar mass of X is 30 g/mol</p><p></p><p>To Find :-</p><p></p><p>Molar mass of Y</p><p></p><p>Solution :-</p><p></p><p>We will use the following formula here to solve this one :-</p><p></p><p>\implies \boxed{ \sf \: \dfrac{T_{x} }{T_{y}} = \sqrt{ \dfrac{M_{x}}{M_{y}} } }⟹TyTx=MyMx</p><p></p><p>where,</p><p></p><p>Tx = Time taken by X gas to be diffused through the holeTy = Time taken by Y gas to be diffused through the holeMx = Molar mass of X gasMy = Molar mass of Y gas</p><p></p><p>➦ Now, substituting the values,</p><p></p><p>\implies \boxed{ \sf \: \dfrac{15 }{25} = \sqrt{ \dfrac{30}{M_{y}} } }⟹2515=My30</p><p></p><p>\implies \boxed{ \sf \: \dfrac{3 }{5} = \sqrt{ \dfrac{30}{M_{y}} } }⟹53=My30</p><p></p><p>➦ Squaring both sides,</p><p></p><p>\implies \boxed{ \sf \: \dfrac{9 }{25} = { \dfrac{30}{M_{y}} } }⟹259=My30</p><p></p><p>\implies \sf \: M_{y} = \dfrac{25 \times 30}{9}⟹My=925×30</p><p></p><p>\implies \sf \boxed{ \red{ \bold{ M_{y} = 83.33}}}⟹My=83.33</p><p></p><p>$