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
Answer:
mass of HNO₃ = 0.378 g
Explanation:
Normality = Molarity * number of equivalents
Molarity = Normality/number of equivalents
normality of HNO₃ = 0.30 N, Volume = 20 mL
HNO₃ ionizes in the following way:
HNO₃(aq) ----> H⁺ + NO₃⁻
Therefore, number of equivalents for HNO₃ is 1
molarity of HNO₃ = 0.30/1 =0.30 mol/dm³
Using the formula, molarity = number of moles/volume in liters
number of moles = molarity * volume
Number of moles of HNO₃ = 0.30 mol/dm³ * 20ml * 1 dm³ /1000 mL
number of moles = 0.006 moles
From the formula, mass = number of moles * molar mass
molar mass of HNO₃ = 63.0 g/mol
mass = 0.006 * 63
mass of HNO₃ = 0.378 g
Answer:
<h2>A. Mercury's orbit is shorter than Earth</h2>
M = 2 . 8 . 2
Valence Electron of M = 2
M ==> M⁺² + 2 e⁻
a. M⁺² + OH⁻ ==> M(OH)₂
b. M⁺² + PO₄⁻³ ==> M₃(PO₄)₂
