Question

Which of the solution has the highest freezing point?
a) 0.25 m CH3OH
b) 0.05 m K2SO4
c) 0.1 m NaCl
d) 0.1 m MgBr2

What is the molar solubility (in M) of helium at 1.0 atm and 25 °C?
a) 0.037
b) 0.00037
c) 0.37
d) 0.0037

Answer 1

Freezing point depression = freezing point of solvent- freezing point of solution = kf*m*i

Freezing point of solution = 0 - kf*m*I

lower Kf*m*I gives higher freezing point

i=van't Hoff factor 1 for CH3OH, 4 for K2SO4, 2 for NaCl and 3 for MgBr2

CH3OH= 0.25kf for K2SO4 , it is = 0.05*4i =0.2i and NaCl =0.2i and MgBr2= 3i

lower kfmi value is there for NaCL and K2SO4. they will have high freezing point

b) From PV= nRT

P= 1 atm T= 25+273.15 =298.15   R=0.08206 L.atm/mole.K

n/V=P/RT = 1/ (25+273.15)*0.08206

=0.0408 moles/L