I think a controlled experiment
Answer is: mass of salt is 311,15 g.
V(H₂O) = 1,48 l · 1000 ml/l = 1480 ml.
m(H₂O) = 1480 g = 1,48 kg.
d(solution) = 1,00 g/ml.
ΔT(solution) = 13,4°C = 13,4 K.
Kf = 1,86 K·kg/mol; cryoscopic constant of water
i(NaCl) = 2; Van 't Hoff factor.
ΔT(solution) = Kf · b · i.
b(NaCl) = 13,4 K ÷ (1,86 K·kg/mol · 2).
b(NaCl) = 3,6 mol/kg.
n(NaCl) = 3,6 mol · 1,48 kg= 5,328 mol.
m(NaCl) = 5,328 mol · 58,4 g/mol = 311,15 g.
All Elements are obtained in the periodic table & water is Not an element it is a compound.
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
<h3>What is the boiling-point elevation?</h3>
Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.
- Step 1: Calculate the molality of the solution.
We will use the definition of molality.
b = mass solute / molar mass solute × kg solvent
b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m
- Step 2: Calculate the boiling-point elevation.
We will use the following expression.
ΔT = Kb × m × i
ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C
where
- ΔT is the boiling-point elevation
- Kb is the ebullioscopic constant.
- b is the molality.
- i is the Van't Hoff factor (i = 2 for NaCl).
The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:
100 °C + 0.140 °C = 100.14 °C
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
Learn more about boiling-point elevation here: brainly.com/question/4206205
