One mole of glucose is dissolved in $2 \, moles$ of water. The vapour pressure of the solution relative to that of water is

The vapour pressure lowering caused by the addition of $100 \, g$ of sucrose (molecular mass $= 342$) to $1000 \, g$ of water if the vapour pressure of pure water at $25 \, ^oC$ is $23.8 \, mm \, Hg$.

The weight in grams of a non-volatile solute (mol. wt. $60$) to be dissolved in $90 \ g$ of water to produce a relative lowering of vapour pressure of $0.02$ is

An aqueous solution of $2\%$ non-volatile solute exerts a pressure of $1.004 \, bar$ at the normal boiling point of the solvent. The molar mass of the solute is $......... \, g/mol$.

The vapour pressure of water at $23^{\circ} C$ is $19.8 \ mm$. If $0.1 \ mole$ of glucose is dissolved in $178.2 \ g$ of water,what is the vapour pressure (in $mm$) of the resultant solution?

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?