The vapour pressures of two volatile liquids $A$ and $B$ at $25^{\circ} C$ are $50 \ Torr$ and $100 \ Torr,$ respectively. If the liquid mixture contains $0.3$ mole fraction of $A$,then the mole fraction of liquid $B$ in the vapour phase is $\frac{x}{17}$. The value of $x$ is $...$
- A$140$
- B$89$
- C$14$
- D$54$
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The vapour pressure of water at $20^\circ C$ is $17.5 \ mm \ Hg$. If $18 \ g$ of glucose $(C_6H_{12}O_6)$ is added to $178.2 \ g$ of water at $20^\circ C$,what will be the vapour pressure of the resulting solution (in $mm \ Hg$)?
Medium
View SolutionWhat is the vapour pressure of a solution containing $1.8 \ g$ of glucose in $16.2 \ g$ of water,if the vapour pressure of pure water is $32 \ mm \ Hg$ (in $mm \ Hg$)?
Which of the following changes will cause an increase in the vapour pressure of a $1 \ m$ aqueous $KI$ solution at the same temperature?
$2\, \text{moles}$ of liquid $A$ is mixed with $3\, \text{moles}$ of liquid $B$. Calculate the mole fraction of $A$ in the vapour phase if $P_A^o = 100\, \text{mm}$ and $P_B^o = 200\, \text{mm}$.
Medium
View Solution$1 \ mol$ each of the following solutes are taken in $5 \ mol$ of water:
$(i) \ NaCl$
$(ii) \ K_2SO_4$
$(iii) \ Na_3PO_4$
$(iv) \ \text{glucose}$
Assuming $100 \%$ ionization of the electrolytes,the relative decrease in vapor pressure will be in the order:
$(i) \ NaCl$
$(ii) \ K_2SO_4$
$(iii) \ Na_3PO_4$
$(iv) \ \text{glucose}$
Assuming $100 \%$ ionization of the electrolytes,the relative decrease in vapor pressure will be in the order:
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