A gas diffuse $$\dfrac{1}{5}$$ times as fast as hydrogen. Its molecular weight is:
Let,
$$R_{H_2}$$ be $$x$$
$$R_{G}$$ be $$\dfrac{1}{5}x$$
applying Graham's law of diffusion
$$\dfrac{R_{H_2}}{R_G}=\sqrt{\dfrac{M_G}{M_{H_2}}}$$
Where $$R_G$$ &$$M_G$$ represents the rate of diffusion and molecular mass of the gas.
Putting the values in the above equation
$$\dfrac{x}{\dfrac{x}{5}}=\sqrt{\dfrac{M_G}{2}}$$
$$5=\sqrt{\dfrac{M_G}{2}}$$
Squaring both sides
$$5^2=\dfrac{M_G}{2}$$
$$M_G=25\times 2=50$$
According to Graham's law, at a given temperature the ratio of the rates of diffusion $$\dfrac {r_A}{r_B}$$ of gases $$A$$ and $$B$$ is given by: (where $$p$$ and $$M$$ are pressures and molecular weights of gases $$A$$ and $$B$$ respectively)
Define Graham’s law of diffusion. Give its mathematical formulation.
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The gases $$A$$ and $$B$$ are kept in two cylinders and allowed to diffuse through a small hole. If the time taken by the gas $$A$$ is four times the time taken by gas $$B$$ for diffusion of equal volume of gas under similar conditions of temperature and pressure, what will be the ratio of their molecular weights?
An open flask containing air is heated from 300 K to 500 K . What percentage of air will be escaped to the atmosphere, if pressure is keeping constant?
According to Graham's law, at a given temperature the ratio of the rates of diffusion $$\dfrac {r_A}{r_B}$$ of gases $$A$$ and $$B$$ is given by: (where $$p$$ and $$M$$ are pressures and molecular weights of gases $$A$$ and $$B$$ respectively)
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If some moles of $$O_2$$ diffuse in $$18$$ sec and same moles of an unknown gas diffuse in $$45$$ sec then what is the molecular weight of the unknown gas?
Rate of diffusion of a gas is