Question

Boiling point of a solvent is 80.2°



c. When 0.419 g of the solute of molar mass 252.4 g mol-1, is dissolved in 75 g of above solvent, the boiling point of solution is found to be 80.26°



c. Find molal elevation constant.

Answer 1

252.4g + 75g

= 327.4g

= 0.419 + 327.4g

= 327.819g

80.26℃

this is a latent heat of fusion

Answer 2

Molal elevation constant is $2.7^0C/m$  

Explanation:

Elevation in boiling point:

$T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = $80.26^oC$

$T^o_b$ = boiling point of solvent = $80.2^oC$

$k_b$ = boiling point constant  = ?

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

$w_2$ = mass of solute = 0.419 g

$w_1$ = mass of solvent = 75 g

$M_2$ = molar mass of solute= 252.4 g/mol

Now put all the given values in the above formula, we get:

$(80.26-80.2)^oC=1\times k_b\times \frac{(0.419g)\times 1000}{252.4\times (75g)}$

$0.06^oC=1\times k_b\times 0.022$

$k_b=2.7^0C/m$

Therefore, the molal elevation constant is $2.7^0C/m$  

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