What weight of glucose must be dissolved in $100 \ g$ of water to lower the vapour pressure by $0.20 \ mm \ Hg$ (in $g$)? (Assume dilute solution is being formed) Given: Vapour pressure of pure water is $54.2 \ mm \ Hg$ at room temperature. Molar mass of glucose is $180 \ g \ mol^{-1}$

An aqueous solution is $1.00 \ molal$ in $KI$. Which change will cause the vapour pressure of the solution to increase?

How many grams of solute (molecular weight $60$) are required to be dissolved in $180 \ g$ of water to reduce the vapour pressure to $\frac{5}{6}$ of the pure water?

Two different solutions were prepared by dissolving different quantities of glucose in the same quantity of water. For the given graph,which statement is incorrect?

Vapour pressure of chloroform $(CHCl_3)$ and dichloromethane $(CH_2Cl_2)$ at $25^\circ C$ are $200 \ mm \ Hg$ and $41.5 \ mm \ Hg$ respectively. Vapour pressure of the solution obtained by mixing $25.5 \ g$ of $CHCl_3$ and $40 \ g$ of $CH_2Cl_2$ at the same temperature will be (Molecular mass of $CHCl_3 = 119.5 \ u$ and molecular mass of $CH_2Cl_2 = 85 \ u$)

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