Question

B.Calculate the boiling point and freezing point of the following solutions:

1.0.575 molal aqueous solution

2.10 g of C6 H12 O6 dissolved in 259 g of water

3.0.5 mole Br2 dissolved in 507 g chloroform

D.A solution of propylene glycol (antifreeze) in the car’s radiator has a freezing point of -20 degree celcius.What is the molarity of the solution?

E.The boiling point of an aqueous solution is 102.5 degree celcius. What is the freezing point of the solution?

 

Answer 1

B.

Taking $k_b$ for water $=0.515°C/M$

and $k_f$ for water $=1.86°C/M$

1.

$\Delta\>T_b=0.515×0.575=0.30$

Boiling point $=100+0.3=100.3°C$

$\Delta\>T_f=1.86×0.575=1.1$

Freezing point $=0-1.1=-1.1°C$

2.

Molar mass of C₆H₁₂O₆ $=180g/mol$

Concentration $=\frac{1000}{259}×\frac{10}{180}=0.2145M$

$\Delta\>T_b=0.2145×1.86=0.40$

Freezing point $=0-0.40=-0.40$