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Standard XII

Chemistry

Stoichiometric Calculations

The density o...

Question

The density of the vapour of a substance at 1 $a t m$ pressure and 500 $K$ is 0.36 $k g$ $m^{- 3}$ . The vapour effuses through a small hole at a rate of 1.33 times faster than oxygen under the same condition. Determine the molar mass in the form of x + 10. The value of x is _____.

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Solution

$\frac{r_{(v)}}{r (O_{2})} = \sqrt{\frac{M_{(O_{2})}}{M_{(v)}}}$

$∴ 1.33 = \sqrt{\frac{32}{M_{(v)}}}$

$M_{(v)} = 18.1 g m o l^{- 1}$

Therefore, the integer value is 18.
$10 + x = 18$

$x = 8$

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The density of the vapour of a substance at 1 atm pressure and 500 K is 0.36 kg m−3. The vapour effuses through a small hole at a rate of 1.33 times faster than oxygen under the same condition. Determine the molar mass in the form of x + 10. The value of x is _____.

r(v)r(O2)=√M(O2)M(v)∴1.33=√32M(v)M(v)=18.1gmol−1Therefore, the integer value is 18.10+x=18x=8

The density of the vapour of a substance at 1 $a t m$ pressure and 500 $K$ is 0.36 $k g$ $m^{- 3}$ . The vapour effuses through a small hole at a rate of 1.33 times faster than oxygen under the same condition. Determine the molar mass in the form of x + 10. The value of x is _____.

$\frac{r_{(v)}}{r (O_{2})} = \sqrt{\frac{M_{(O_{2})}}{M_{(v)}}}$

$∴ 1.33 = \sqrt{\frac{32}{M_{(v)}}}$

$M_{(v)} = 18.1 g m o l^{- 1}$

Therefore, the integer value is 18.
$10 + x = 18$

$x = 8$