Question

Doblydorlog on to com 2. Rank order the following aqueous solutions in order of increasing boiling point elevation. 0.100 mol/kg HBr 3.1.2 5.2 ЈТве - Комі 0.100 mol/kg MgSO4 =. 0.010 mol/kg NaCl lowing 8.00 a 0.010 mol/kg glucose 0.010 mol/kg CaCl2 A. HBr <MgSO4 = CaCl2<NaCl < glucose B. MgSO4 < HBr < CaCl2 < NaCl <glucose C.) glucose < NaCl <CaCl2 <HBr <MgSO4 D. glucose < CaCl2< NaCl <HBr <MgSO4 E. glucose < CaCl2< NaCl <MgSO4 = HBr 0.001) astonbod e nbora

Answer 1

Boiling point elevation is an example for the colligative properties. It is the phenomenon in which the boiling point of a liquid is increased on the addition of a compound.

The elevation in boiling point is given by the equation,

ΔΤ = iK5 m

Here, $\bigtriangleup T$ is the increase in boiling point

is the van't Hoff factor

$K_{b}$ is the boiling point elevation constant

is the molality of the solution

Thus from the equation it can be identified that the elevation in boiling point is directly proportional to van't Hoff factor and molality of the solution.

van't Hoff factor is defined as the ratio of number of particles in the solution after the dissociation to the number of particles in the solution before dissociation.

For non electrolytes, $i=1$

For strong electrolytes,

For weak electrolytes,

The value for each of the given electrolytes are,

$MgSO_{4}=2(MgSO_{4}\rightarrow Mg^{2+}+SO_{4}^{2-})$

$HBr=2(HBr\rightarrow H^{+}+Br^{-})$

$NaCl=2(NaCl\rightarrow Na^{+}+Cl^{-})$

$CaCl_{2}=3(CaCl_{2}\rightarrow Ca^{2+}+2Cl^{-})$

On comparing the molality values,

MgSO4, Ta2 x 0.100m = 0.200m

HBr = 2, AT a2 x 0.100m = 0.200m

$NaCl=2,\bigtriangleup T\alpha 2\times 0.010m=0.020m$

$CaCl_{2}=3, \bigtriangleup T\alpha 3\times 0.010m=0.030m$

$Glucose=1,\bigtriangleup T\alpha 1\times 0.010m=0.010m$

Thus the order of increasing boiling point of the electrolytes are given by,

Glucose<NaCl<CaCl2<HBr<MgSO4