the missing options are :
(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point
answer : the correct answer is option (3) higher boiling point and a lower freezing point
explanation :
when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.
let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m
step 1 : find molality of the solution
let us assume we have 1 l of solution.
![m = \frac{moles (solute)}{l ( solution)}](/tex.php?f= m = \frac{moles (solute)}{l ( solution)} )
![1 m = \frac{moles (cacl_{2})}{1 l }= 1 mol cacl_{2}](/tex.php?f= 1 m = \frac{moles (cacl_{2})}{1 l }= 1 mol cacl_{2} )
let us convert this to grams of cacl₂.
molar mass of cacl2 is 111 g/mol
grams of cacl₂ = ![1 mol \times \frac{111g}{mol} = 111 g cacl_{2}](/tex.php?f= 1 mol \times \frac{111g}{mol} = 111 g cacl_{2} )
mass of solute = 111 g
assuming density of solution same as water which is 1g/ml we have ,
mass of solution = density * volume
mass of solution = ![\frac{1g}{ml} \times 1000 ml = 1000 g](/tex.php?f= \frac{1g}{ml} \times 1000 ml = 1000 g )
mass of solution = mass of solvent + mass of solute
mass of solute = 1000g - 111g
mass of solute 889 g = 0.889 kg
the molality of the solution can be calculated as
![molality = \frac{mol (solute)}{kg ( solvent)}](/tex.php?f= molality = \frac{mol (solute)}{kg ( solvent)} )
![molality = \frac{1mol}{0.889kg}](/tex.php?f= molality = \frac{1mol}{0.889kg} )
molality = 1.12 m
step 2 : find boiling point elevation.
the elevation in boiling point is calculated as
![\bigtriangleup t_{b} = i \times k_{b} \times m](/tex.php?f= \bigtriangleup t_{b} = i \times k_{b} \times m )
here δtb = boiling point elevation
i = vant hoff factor which is the number of ions formed by the solute .
cacl₂ dissociates as ![cacl_{2} \rightarrow ca^{2+} + 2 cl^{-}](/tex.php?f= cacl_{2} \rightarrow ca^{2+} + 2 cl^{-} )
it forms 3 ions . therefore,
i = 3
m = molality which is 1.12 m
kb is constant for a given solvent. for water it is 0.512
![\bigtriangleup t_{b} = 3 \times 0.512 \times 1.12](/tex.php?f= \bigtriangleup t_{b} = 3 \times 0.512 \times 1.12 )
![\bigtriangleup t_{b} = 1.7](/tex.php?f= \bigtriangleup t_{b} = 1.7 )
the boiling point of water increases by 1.7 °c
step 3: find freezing point depression.
when a non volatile solute is added to water, it results in decrease in freezing point.
the formula to find freezing point depression is
![\bigtriangleup t_{f} = i \times k_{f} \times m](/tex.php?f= \bigtriangleup t_{f} = i \times k_{f} \times m )
kf is constant for water and its value is 1.86
![\bigtriangleup t_{f} = 3 \times 1.86 \times 1.12](/tex.php?f= \bigtriangleup t_{f} = 3 \times 1.86 \times 1.12 )
![\bigtriangleup t_{f} = 6.2](/tex.php?f= \bigtriangleup t_{f} = 6.2 )
the freezing point of water decreases by 6.2 °c
therefore we can say that, the solution will have a higher boiling point and a lower freezing point