The partial vapour pressure of any volatile component of a solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution is called:

$3 \ g$ of urea is dissolved in $45 \ g$ of water. The relative lowering of vapour pressure is

Two open beakers,one containing a solvent and the other containing a mixture of that solvent with a non-volatile solute,are together sealed in a container. Over time:

The vapor pressure of water at $300 \, K$ is $12.3 \, kPa$. Calculate the vapor pressure of a $1 \, molal$ solution containing a non-volatile solute. (in $, kPa$)

$1 \ mol$ each of the following solutes are taken in $5 \ mol$ of water:
$(i) \ NaCl$
$(ii) \ K_2SO_4$
$(iii) \ Na_3PO_4$
$(iv) \ \text{glucose}$
Assuming $100 \%$ ionization of the electrolytes,the relative decrease in vapor pressure will be in the order:

One mole of a non-volatile solute is dissolved in two moles of water. What is the vapor pressure of this solution relative to that of pure water?