The mass of KNO₃ that is required to be added to 275 ml of water is; 108.3 g
<h3>Dissolution of a Solute</h3>
We are given;
K f = 1.86 °C.kg/m
Mass of solution = 275 g = 0.275 Kg
Freezing point of solution = −14.5 °C
Freezing point of pure water = 0.0 °C
We know that;
Δ T = Freezing point of solve nt - Freezing point of solut ion
Thus;
Δ T = 0°C - ( -14.5 °C)
Δ T = 14.5 °C
Also;
Δ T = K f * m * i
where;
K f is free zing con stant
m is m o l a l i t y
i is V a n t H o f f factor
Since KNO₃ produces two particles then; i = 2
Making the m the subject of the formula gives;
m = ΔT /(K × i)
m = 14.5 °C/(1.86 °C . kg/m o l × 2)
m = 3.898 m
Now;
M o l a l i t y = number of mo les/mass of solution in kilograms
Thus;
number of mo les of solute = m o l a l i t y × mass of solution in kilograms
number of mo les of solute= 3.898 m × 0.275 Kg
number of mo les of solute= 1.072 moles
Thus;
Number of mo les = mass/mo lar mass
M o lar mass of KNO₃ = 101 g/mol
Mass = Number of mo les × 101 g/mol
Mass = 1.072 moles × 101 g/mol
Mass of KNO₃ = 108.3 g
Read more about dissolution of a solute at; brainly.com/question/1945924
