This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{Rate_{X}}{Rate_{C_3H_8}}=1.55$

$\frac{Rate_{X}}{Rate_{C_3H_8}}=\sqrt{\frac{M_{C_3H_8}}{M_{X}}}$

$1.55=\sqrt{\frac{44}{M_{X}}$

Squaring both sides and solving for $M_{X}$

$M_{X}=18.3g/mol$

Hence, the molar mas of unknown gas is 18.3 g/mol.

5 0

3 years ago

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Determine the mass of Al(C2H3O2)3 that contains 2.63 Χ 1024 atoms of oxygen.

sergejj [24]

The molecular formula of compound is $Al(C_{2}H_{3}O_{2})_{3}$ .

Since, in 1 mole of $Al(C_{2}H_{3}O_{2})_{3}$ there are $6.023\times 10^{23}$ atoms of $Al(C_{2}H_{3}O_{2})_{3}$ .

Thus, according to molecular formula, in 1 mole of $Al(C_{2}H_{3}O_{2})_{3}$ there are $6\times 6.023\times 10^{23}=3.61\times 10^{24}$ atoms of oxygen atoms.

1 atom of oxygen will be present in $\frac{1}{3.61\times 10^{24}}$ moles of $Al(C_{2}H_{3}O_{2})_{3}$ . Thus,

$2.63\times 10^{24} atoms of oxygen \rightarrow \frac{2.63\times 10^{23}}{3.61\times 10^{24}}= 0.7285 moles of Al(C_{2}H_{3}O_{2})_{3}$ .

Molar mass of $Al(C_{2}H_{3}O_{2})_{3}$ is 236 g/mol, mass can be calculated as follows:

$m=n\times M=0.7285 mol\times 236 g/mol=172 g$

Therefore, mass of $Al(C_{2}H_{3}O_{2})_{3}$ will be 172 g.

5 0

3 years ago

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