$12 \ g$ of a nonvolatile solute dissolved in $108 \ g$ of water produces the relative lowering of vapour pressure of $0.1$. The molecular mass of the solute is

$18 \ g$ glucose $(C_6H_{12}O_6)$ is added to $178.2 \ g$ water. The vapour pressure of this aqueous solution at $100 \ ^{\circ}C$ (in $torr$) is:

The vapour pressure of acetone at $20\,^{\circ}C$ is $185\,torr.$ When $1.2\,g$ of a non-volatile substance was dissolved in $100\,g$ of acetone at $20\,^{\circ}C,$ its vapour pressure was $183\,torr.$ The molar mass $(g\,mol^{-1})$ of the substance is:

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 $20\,^{\circ}C,$ the vapour pressure of the resulting solution will be $.........\, mm$ of $Hg$.

$5 \ cm^3$ of acetone is added to $100 \ cm^3$ of water. The vapour pressure of water over the solution:

Find the molar mass of solute when $2 \ g$ is dissolved in $60 \ g$ of benzene,and the relative lowering of vapour pressure is $0.06$. (Molar mass of benzene is $78 \ g \ mol^{-1}$)