An osmolarity of saline solution is 308 mosmol/L.
m(NaCl) = 9 g; the mass of sodium chloride
V(solution) = 1 L; the volume of the saline solution
n(NaCl) = 9 g ÷ 58.44 g/mol
n(NaCl) = 0.155 mol; the amount of sodium chloride
number of ions = 2
Osmotic concentration (osmolarity) is a measure of how many osmoles of particles of solute it contains per liter.
The osmolarity = n(NaCl) ÷ V(solution) × 2
The osmolarity = 0.154 mol ÷ 1 L × 2
The osmolarity = 0.154 mol/L × 1000 mmol/m × 2
The osmolarity of the saline solution = 308 mosm/L.
More about osmolarity: brainly.com/question/13258879
#SPJ4
To calculate the atomic mass of a single atom of an element, add up the mass of protons and neutrons.
Chemical property of the substance but then again it could be physical
Answer:
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
Explanation:
One colligative property is the freezing point depression due the addition of a solute. The equation is:
ΔT=Kf*m*i
<em>Where ΔT is change in temperature = 0.400°C</em>
<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>
<em>m is molality of the solution (Moles of solute / kg of solvent)</em>
<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>
Replacing:
0.400°C =1.86°C/m*m*1
0.400°C / 1.86°C/m*1 = 0.215m
As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:
0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.
The mass of ethylene glycol must be added is:
0.0602 moles * (62.10g / mol) =
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
<em />
