Solutions - Class 12 Medical Chemistry - Extra Questions

Higher is the _____ (mass/volume) of solute in a solution lower is its vapour pressure.

The relative lowering of the vapor pressure of a dilute solution is directly proportional to the mole fraction of the solute.

Higher is the mass of solute in a solution, higher will be its mole fraction, higher will be the relative lowering in the vapor pressure and lower will be the vapor pressure of the solution.

So, the correct answer is mass.


A non volatile solute ________ (decrease/increase) the rate of evaporation of the solvent and thereby lowers the vapour pressure of solvent.

In case of pure water, the surface area is completely occupied by water molecules. In case of solution, the surface area is partly occupied by water molecules and partly occupied by non volatile solute. 

Hence, the escaping tendency of water molecules from pure water is greater than the escaping tendency of water molecules from a solution because of more surface area. Thus, the rate of evaporation of the solution is less than the rate of evaporation of water.

So, when a non-volatile solute is added to water, its vapor pressure is lowered.


The vapour pressure of ethanol and methanol are $$44.5\ mm$$ and $$88.7\ mm\ Hg$$ respectively. An ideal solution is prepared at the same temperature by mixing $$60\ g$$ of ethanol with $$40\ g$$ of methanol. Calculate total vapour pressure of the solution.


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Sugar and Salt solutions are________.

True solutions 


$$10^{-3}$$ mol of $$CuSO_{4}\cdot 5H_{2}O$$ is introduced in a $$1.9\ L$$ vessel maintained at a constant temperature of $$27^{\circ}C$$ containing moist air at relative humidity of $$12.5$$%. What is the final molar composition of solid mixture?
For $$CuSO_{4}.5H_{2}O(s)\rightleftharpoons CuSO_{4}(s) + 5H_{2}O(g), K_{p}(atm) = 10^{-10}$$. Take vapor pressure of water at $$27^{\circ}C$$ as $$28$$ torrs.

$${ CuSO }_{ 4 }.5{ H }_{ 2 }O\left( s \right) \rightleftharpoons { CuSo }_{ 4 }\left( s \right) +5{ H }_{ 2 }O\left( g \right) $$
                                                           $${ K }_{ P }={ 10 }^{ -10 }$$
$${ K }_{ P }={ P }_{ { H }_{ 2 }O }^{ 5 },\quad \therefore \quad { P }_{ { H }_{ 2 }O }^{ 5 }={ 10 }^{ -10 }\Rightarrow { P }_{ { H }_{ 2 }O }={ 10 }^{ -12 }atm$$
$$\therefore \quad { n }_{ { H }_{ 2 }O }=\dfrac { 0.0368atm\times 1.9L }{ 0.0821lit-atm\times { mol }^{ -1 }{ K }^{ -1 }\times 300 } =0.003$$
$$\therefore \quad { n }_{ { CuSO }_{ 4 } }=0.003/5=0.0006$$ moles
$${ n }_{ { CuSO }_{ 4 }.{ 5H }_{ 2 }O }={ 10 }^{ -13 }-0.0006=0.0004$$ moles
$$\therefore$$  The final mixture contains $$0.0004$$ moles and $$0.0006$$ moles of $${ CuSO }_{ 4 }.{ 5H }_{ 2 }O\left( s \right) $$ and $${ CuSO }_{ 4 }\left( s \right) $$ respectively.


Class 12 Medical Chemistry Extra Questions