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An evacuated glass vessel weighs 40.0 \mathrm{~g} when empty, 135.0 \mathrm{~g} when filled with a liquid of density 0.95 \mathrm{~g} \mathrm{} \mathrm{~mL}^{-1}$ and $40.5 \mathrm{~g} when filled with an ideal gas at 0.82 \mathrm{~atm}$ at $250 \mathrm{~K}. The molar mass of the gas in \mathrm{g} \mathrm{~mol}^{-1} is :
(Given : \mathrm{R}=0.082 \mathrm{~L}$ atm $\mathrm{K}^{-1} \mathrm{~mol}^{-1} )

Option: 1

35


Option: 2

50


Option: 3

75


Option: 4

125


Answers (1)

best_answer

Given, mass of empy glass vessel  \mathrm{= 40\, g}
Total mass \mathrm{= 135\, g}.

\mathrm{\text{density of liquid }=0.95 \mathrm{~g}\, \mathrm{ml^{-1}}}
\mathrm{R=0.0 8 2\, L \: \: atm \mathrm{~K}^{-}~mol ^{-1}}


\mathrm{\text{mass of Liquid} =135-40=95 \mathrm{~g}.}

\mathrm{\text { Volume of Liquid }=\frac{\text { mass }}{ \text { density }} =\frac{95 \mathrm{~g}}{0.95 \mathrm{~g} / \mathrm{ml}}=100\, \mathrm{ml}}
                                                                                                \mathrm{= 0.1\, L}

Now, for ideal gas-

\mathrm{\text{mass of ideal gas} =40.5-40=0.5 \mathrm{~g}}

From ideal gas equation 
\mathrm{P V=n R T}

Putting values -
\mathrm{0.82\operatorname{atm} \times 0.1 \mathrm{~L}=\left ( \frac{0.5}{M} \right ) mol \times 0.082\, \mathrm{L\: atm\: K}^{-1} \mathrm{~mol}^{-1} \times 250\, K}
Solving equation, we get -
\mathrm{M=125\, \mathrm{g\: mol}^{-1}}

Option (4) is correct

Posted by

Rakesh

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