$A$ solution of two miscible liquids showing negative deviation from Raoult's law will have :

At $373 \ K$,heptane and octane form an ideal solution. The vapor pressures of the two pure liquid components (heptane and octane) are $105 \ kPa$ and $45 \ kPa$,respectively. The vapor pressure of the solution formed by mixing $25 \ g$ of heptane and $35 \ g$ of octane is ........ $kPa$. (Molar masses of heptane and octane are $100 \ g \ mol^{-1}$ and $114 \ g \ mol^{-1}$,respectively.)

At $20^o C$,benzene and toluene form an ideal solution. The pure vapor pressure of benzene is $75 \ torr$ and that of toluene is $22 \ torr$. What is the partial vapor pressure of benzene in a solution containing $78 \ g$ of benzene and $46 \ g$ of toluene at $20^o C$?

What are the values of $\Delta H_{mix}$ and $\Delta V_{mix}$ for an ideal solution?

At $T \ K$,two liquids $A$ and $B$ form an ideal solution. The vapour pressures of pure liquids $A$ and $B$ at that temperature are $400 \ mm \ Hg$ and $600 \ mm \ Hg$ respectively. If the mole fraction of liquid $B$ is $0.3$ in the mixture,the mole fractions of $A$ and $B$ in the vapour phase respectively are

Which of the following will form an ideal solution?