Question

A soln of 12.5g of an unknown solute in 170 g of water gave a boiling point elevation of 0.63k calculate molar mass of solute

Answer 1

molecular mass of solute =61.8 gm

ΔTb =kb molality

0.63 =

0.53 x12.5/moleular mass x1000/170

MM = 61.8 gm

Answer 2

Molar mass of solute is 60 g/mol

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,

$T_b-T^o_b=0.63K$

$k_b$ = boiling point constant  of water = $0.512K/m$

m = molality

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

$w_2$ = mass of solute = 12.5 g

$w_1$ = mass of solvent (water) = 170 g

$M_2$ = molar mass of solute  = ?

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

$0.63=1\times (0.512^oC/m)\times \frac{(12.5g)\times 1000}{M_2\times (170g)}$

$M_2=60g/mol$

Therefore, the molar mass of solute is 60 g/mol

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