Solution us here,
let the rate of diffusion of H2(r1) be x and⁷ the rate of diffusion of gas(r2) be x+x/6=7x/6.
Molar mass of H2 (M1)=2 g
Molar mass of gas(M2)= ?
Now, from the Ghram's law of diffusion of gas,

here, it looks like I have done wrong.
But all of the answers in option are wrong.
Because, rate of diffusion of any gas is inversely proportional to the square root of its malor mass.
And in the question, gas has higher rate of diffusion than hrdrogen. So it should have molar mass less than hydrogen.
Go through your work carefully i see a mistake in number 2
Answer:
364 K or 91°C
Explanation:
Applying,
V₁/T₁ = V₂/T₂................ Equation 1
Where V₁ = Initial Volume, V₂ = Final volume, T₁ = initial Temperature, T₂ = final Temperature.
make T₂ the subject of the equation,
T₂ = V₂T₁/V₁................. Equation 2
From the question,
Given: V₁ = 375 mL, V₂ = 500 mL, T₁ = 0.0°C = (273+0) K = 273 K
Substitute these values into equation 2
T₂ = (500×273)/375
T₂ = 364 K
T₂ = (364-273) °C = 91 °C
Here we have to get the residual concentration of Cu²⁺ in the reaction.
The residual concentration of [Cu²⁺] is 2.137×10⁸ m
The reaction between ammonia (NH₃) and CuSO₄ in ionic form is - Cu²⁺ + 4 NH₃ = [Cu(NH₃)₄]²⁺.
Thus 1 mole of Cu²⁺ will react with 4 moles of NH₃.
Now 6m NH₃ or 6 moles of ammonia will react with
= 1.5 moles of Cu²⁺.
Thus the Nernst equation of the reaction is
= E° +
log
.
Here the number of moles is 3, as 6 moles of ammonia will react with 1.5 moles of cu²⁺. and the standard reduction potential (E°) of Cu²⁺/Cu system is (+) 0.674V and the cell potential is 0.92V
On plugging the values in the equation, we get,
0.92 = 0.674 +
log[Cu²⁺]
Or, 0.029 log [Cu²⁺]= 0.246
Or, log [Cu²⁺] = 8.33
Or, [Cu²⁺] = 2.137×10⁸