JEE MAIN - Chemistry (2018 - 16th April Morning Slot - No. 21)
The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol-1 ) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75%, is :
37.5 g
75 g
150 g
50 g
Explanation
Molar mass of octane
(C8 H18) = 8 $$ \times $$ 12 + 18 = 114 g/mol
Let, $$\omega $$ is the mass of solute.
Relative lowering vapour pressure,
$${{\Delta P} \over P}$$ = $${{{\omega \over {50}}} \over {{\omega \over {50}} + {{114} \over {114}}}}$$
$$ \Rightarrow $$$$\,\,\,\,$$ $${{75} \over {100}}$$ = $${{{\omega \over {50}}} \over {{\omega \over {50}} + 1}}$$
$$ \Rightarrow $$ $$\,\,\,\,$$ $${3 \over 4}$$ $$\left( {{\omega \over {50}} + 1} \right) = {\omega \over {50}}$$
$$ \Rightarrow $$ $$\,\,\,\,$$ $$\omega $$ = 150 g
(C8 H18) = 8 $$ \times $$ 12 + 18 = 114 g/mol
Let, $$\omega $$ is the mass of solute.
Relative lowering vapour pressure,
$${{\Delta P} \over P}$$ = $${{{\omega \over {50}}} \over {{\omega \over {50}} + {{114} \over {114}}}}$$
$$ \Rightarrow $$$$\,\,\,\,$$ $${{75} \over {100}}$$ = $${{{\omega \over {50}}} \over {{\omega \over {50}} + 1}}$$
$$ \Rightarrow $$ $$\,\,\,\,$$ $${3 \over 4}$$ $$\left( {{\omega \over {50}} + 1} \right) = {\omega \over {50}}$$
$$ \Rightarrow $$ $$\,\,\,\,$$ $$\omega $$ = 150 g
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