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Mass of the water : 2.23 g
<h3>Furter explanation</h3>Heat
Q = m.c.Δt
m= mass, g
c = heat capacity, for water : 4.18 J/g° C.
ΔT = temperature
Q= 140 J
Δt = 75 - 60 = 15
mass of the water :
Answer:
The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.
Explanation:
Water potential = Pressure potential + solute potential
We have :
C = 0.15 M, T = 273.15 K
i = 1
The water potential of a solution of 0.15 m sucrose=
(At standard temperature)
The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.