12 gm of glucose is dissolved in 100 gm of water. Its b.p is 100.34°c and the b.p. of water is 100°c. calculate the molar elevation constant of Water?
Answers
Boiling point of a solution is found to be \displaystyle 100.34 \: ^oC \:100.34
o
C. Boiling point of pure
water is \displaystyle 100 \: ^oC \:100
o
C.
The elevation in the boiling point $$ \displaystyle \Delta T_b = 100.34-
100=0.34 \: ^oC$$.
1212 gm glucose (molecular weight 180180 g/mol) is dissolved in 100100 gm water.
The number of moles of glucose $$ \displaystyle = \dfrac {12 \: g}{180 \:
g/mol} = 0.0667 \: mol$$
Mass of water $$ \displaystyle = 100 \: g \times \dfrac {1 \: kg}{1000 \: g}
=0.100 \: kg$$
Molality of solution $$ \displaystyle m = \dfrac {0.0667 \: mol}{0.100 \: kg}
=0.667 \: mol/kg$$
The elevation in the boiling point \displaystyle \Delta T_b = K_b \times m ΔT
b
=K
b
×m
\displaystyle 0.34 \: ^oC= K_b \times 0.667 \: mol/kg 0.34
o
C=K
b
×0.667mol/kg
\displaystyle K_b = 0.51 \: ^oC \: kg/molK
b
=0.51
o
Ckg/mol
The molal elevation constant for water is \displaystyle 0.51 \: ^oC \: kg/mol 0.51
o
Ckg/mol.