Answer:
The molality ( m ) of a solution is the moles of solute divided by the kilograms of solvent. A solution that contains 1.0 mol of NaCl dissolved into 1.0 kg of water is a “one-molal” solution of sodium chloride. The symbol for molality is a lower-case m written in italics.
In order to calculate the molality of a solution divide the moles of solute by the volume of the solution expressed in liters.
I did what I think you meant if it's wrong text me I'll fix it.
Answer:
The volume of the gas at 100°C is 4.6189 liters.
Explanation:
For this problem we are going to use Charles' law. Charles' law states that the volume is directly proportional to temperature given that the pressure is constant. In order to use the equation, the unit of temperature should be in Kelvin.
The working equation is:
=
where V1 and T1 are the initial volume and temperature while V2 and T2 are the final conditions.
Let us convert first the temperatures before solving for the final volume.
To convert Celsius to Kelvin just add 273.15 to temperature in Celsius.
50°C + 273.15 = 323.15 K
100°C + 273.15 = 373.15 K
Solving for the final volume:
V₂ =
V₂ =
V₂ = 4.6189 L
Therefore the final volume of the gas at 100°C is 4.6189 L.
Explanation:
Answer: potassium iodide is the basic test for starch,and the positive test is blue-black coloration, any other test substance which is not starch will give a negative results.
Explanation:
Starch is an example of polysaccharide and since the standard test for it is potassium iodide solution, it gives a positive test.
Diasaccharides e.g maltose are reducing sugars.their standard test is BENEDICT test .
Therefore, in the hydrolysis; starch should give a positve test, while Diasaccharides should give negative rest.
M(dextrose) = 50 g.
V(solution) = 1 L.
n(dextrose) = 50 g ÷ 180 g/mol.
n(dextrose) = 0,27 mol.
Osmotic concentration (osmolarity)<span> is a measure of how many </span><span>osmoles of particles of solute</span><span> it contains </span>per liter.
The osmolarity = n(dextrose) ÷ V(solution).
The osmolarity = 0,27 mol ÷ 1 L.
The osmolarity = 0,27 mol/L · 1000 mmol/m.
The osmolarity (dextrose) = 270 mosm/L.
The osmolarity (dextrose monohydrate) = 50 g÷197 g/mol·1000 =254mosm/L
