277374 $P_{A}$ and $P_{B}$ are the vapour pressures of pure liquid components, $A$ and $B$, respectively of an ideal binary solution. If $\mathrm{x}_{\mathrm{A}}$ represents the mole fraction of component $A$, the total pressure of the solution will be

277375 Vapour pressure of chloroform $\left(\mathrm{CHCl}_{3}\right)$ and dichloromethane $\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right)$ at $25^{\circ} \mathrm{C}$ are $200 \mathrm{~mm}$ $\mathrm{Hg}$ and $41.5 \mathrm{~mm} \mathrm{Hg}$ respectively. Vapour pressure of the solution obtained by mixing $25.5 \mathrm{~g}$ of $\mathrm{CHCl}_{3}$ and $40 \mathrm{~g}$ of $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ at the same temperature will be (Molecular mass of $\mathrm{CHCl}_{3}=\mathbf{1 1 9 . 5} \mathrm{u}$ and molecular mass of $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ $=85 \mathrm{u})$

277376 $K_{f}$ for water is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. If your automobile radiator holds $1.0 \mathrm{~kg}$ of water, then how many grams of ethylene glycol $\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)$ must you add to get the freezing point of the solution lowered to $-2.8^{\circ} \mathrm{C}$ ?

277378 The mass of a non-volatile solute of molar mass $40 \mathrm{~g} \mathrm{~mol}^{-1}$ that should be dissolved in $114 \mathrm{~g}$ of octane to lower its vapour pressure by $20 \%$ is

277374 $P_{A}$ and $P_{B}$ are the vapour pressures of pure liquid components, $A$ and $B$, respectively of an ideal binary solution. If $\mathrm{x}_{\mathrm{A}}$ represents the mole fraction of component $A$, the total pressure of the solution will be

277375 Vapour pressure of chloroform $\left(\mathrm{CHCl}_{3}\right)$ and dichloromethane $\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right)$ at $25^{\circ} \mathrm{C}$ are $200 \mathrm{~mm}$ $\mathrm{Hg}$ and $41.5 \mathrm{~mm} \mathrm{Hg}$ respectively. Vapour pressure of the solution obtained by mixing $25.5 \mathrm{~g}$ of $\mathrm{CHCl}_{3}$ and $40 \mathrm{~g}$ of $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ at the same temperature will be (Molecular mass of $\mathrm{CHCl}_{3}=\mathbf{1 1 9 . 5} \mathrm{u}$ and molecular mass of $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ $=85 \mathrm{u})$

277376 $K_{f}$ for water is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. If your automobile radiator holds $1.0 \mathrm{~kg}$ of water, then how many grams of ethylene glycol $\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)$ must you add to get the freezing point of the solution lowered to $-2.8^{\circ} \mathrm{C}$ ?

277378 The mass of a non-volatile solute of molar mass $40 \mathrm{~g} \mathrm{~mol}^{-1}$ that should be dissolved in $114 \mathrm{~g}$ of octane to lower its vapour pressure by $20 \%$ is

277374 $P_{A}$ and $P_{B}$ are the vapour pressures of pure liquid components, $A$ and $B$, respectively of an ideal binary solution. If $\mathrm{x}_{\mathrm{A}}$ represents the mole fraction of component $A$, the total pressure of the solution will be

277375 Vapour pressure of chloroform $\left(\mathrm{CHCl}_{3}\right)$ and dichloromethane $\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right)$ at $25^{\circ} \mathrm{C}$ are $200 \mathrm{~mm}$ $\mathrm{Hg}$ and $41.5 \mathrm{~mm} \mathrm{Hg}$ respectively. Vapour pressure of the solution obtained by mixing $25.5 \mathrm{~g}$ of $\mathrm{CHCl}_{3}$ and $40 \mathrm{~g}$ of $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ at the same temperature will be (Molecular mass of $\mathrm{CHCl}_{3}=\mathbf{1 1 9 . 5} \mathrm{u}$ and molecular mass of $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ $=85 \mathrm{u})$

277376 $K_{f}$ for water is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. If your automobile radiator holds $1.0 \mathrm{~kg}$ of water, then how many grams of ethylene glycol $\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)$ must you add to get the freezing point of the solution lowered to $-2.8^{\circ} \mathrm{C}$ ?

277378 The mass of a non-volatile solute of molar mass $40 \mathrm{~g} \mathrm{~mol}^{-1}$ that should be dissolved in $114 \mathrm{~g}$ of octane to lower its vapour pressure by $20 \%$ is

277374 $P_{A}$ and $P_{B}$ are the vapour pressures of pure liquid components, $A$ and $B$, respectively of an ideal binary solution. If $\mathrm{x}_{\mathrm{A}}$ represents the mole fraction of component $A$, the total pressure of the solution will be

277375 Vapour pressure of chloroform $\left(\mathrm{CHCl}_{3}\right)$ and dichloromethane $\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right)$ at $25^{\circ} \mathrm{C}$ are $200 \mathrm{~mm}$ $\mathrm{Hg}$ and $41.5 \mathrm{~mm} \mathrm{Hg}$ respectively. Vapour pressure of the solution obtained by mixing $25.5 \mathrm{~g}$ of $\mathrm{CHCl}_{3}$ and $40 \mathrm{~g}$ of $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ at the same temperature will be (Molecular mass of $\mathrm{CHCl}_{3}=\mathbf{1 1 9 . 5} \mathrm{u}$ and molecular mass of $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ $=85 \mathrm{u})$

277376 $K_{f}$ for water is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. If your automobile radiator holds $1.0 \mathrm{~kg}$ of water, then how many grams of ethylene glycol $\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)$ must you add to get the freezing point of the solution lowered to $-2.8^{\circ} \mathrm{C}$ ?

277378 The mass of a non-volatile solute of molar mass $40 \mathrm{~g} \mathrm{~mol}^{-1}$ that should be dissolved in $114 \mathrm{~g}$ of octane to lower its vapour pressure by $20 \%$ is