Question

The lowering of vapour pressure of 0.1M aqueous solutions of Nacl, Cuso4 and k2SO4 will be in the ratio of? Plz elaborate your answer....

Answer 1

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Answer 2

Answer : The lowering of vapor pressure of 0.1 M aqueous solutions of $NaCl,CuSO_4\text{ and }K_2SO_4$ will be in the ratio of $2:2:3$

Explanation :

According to the relative lowering of vapor pressure, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component of the solution multiplied by the vapor pressure of that component in the pure state.

0.1 M means that the 0.1 moles of solute present in 1 liter of solution.

Formula used :

$\frac{\Delta p}{p^o}=i\times X_B$

where,

$p^o$ = vapor pressure of the pure component (water)

$p_s$ = vapor pressure of the solution  

$X_B$ = mole fraction of solute  = 0.1

i = Van't Hoff factor

(a) The dissociation of 0.1 M $NaCl$ will be,

$NaCl\rightarrow Na^++Cl^-$

So, Van't Hoff factor = Number of solute particles = $Na^++Cl^-$ = 1 + 1 = 2

The lowering of vapor pressure of 0.1 M NaCl will be,

$\frac{\Delta p}{p^o}=2\times 0.1=0.2$

(b) The dissociation of 0.1 M $CuSO_4$ will be,

$CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}$

So, Van't Hoff factor = Number of solute particles = $Cu^{2+}+SO_4^{2-}$ = 1 + 1 = 2

The lowering of vapor pressure of 0.1 M $CuSO_4$ will be,

$\frac{\Delta p}{p^o}=2\times 0.1=0.2$

(c) The dissociation of 0.1 M $K_2SO_4$ will be,

$K_2SO_4\rightarrow 2K^{+}+SO_4^{2-}$

So, Van't Hoff factor = Number of solute particles = $2K^{+}+SO_4^{2-}$ = 2 + 1 = 3

The lowering of vapor pressure of 0.1 M $K_2SO_4$ will be,

$\frac{\Delta p}{p^o}=3\times 0.1=0.3$

The ratio of lowering of vapor pressure of 0.1 M solution of $NaCl,CuSO_4\text{ and }K_2SO_4$ will be,

$NaCl:CuSO_4:K_2SO_4$ = 0.2 : 0.2 : 0.3 = 2 : 2 : 3

Therefore, the lowering of vapor pressure of 0.1 M aqueous solutions of $NaCl,CuSO_4\text{ and }K_2SO_4$ will be in the ratio of $2:2:3$