Explain what each quantum number in a quantum number set tells you about the electron. Compare and contrast the locations and pr
2 answers:
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The freezing point depression of a solution containing 30.7 g of glycerin is calculated as -1.65°C
Equating :
It is given that,
Given mass of glycerin is = 30.7 grams (Solute)
Volume of water = 376 mL
or molar -freezing-depression point is = 1.86°C/m
Molar mass of glycerin = 92.09 g/mole
Now, to work out the value, the mass of water should be known. Thus, to calculate, the formula used will be:
Mass = Density X Volume
Mass = 1.0 g/mL X 376 mL
Mass = 376 g or 0.376 Kg
Using the formula of melting point depression, the equation becomes:
Δ = i ×
×m
T⁰- =
in which,
Δ = change in freezing point
Δ = freezing point of solution that has to be find
ΔT° = freezing point of water ()
Since, glycerin is a non-electrolyte, the Van't Hoff factor will be 1.
Substituting the values in the above equation:
0⁰C₋T = 1 ×1.86°C/m ×
= -1.65°C
Thus, the freezing point depression of a solution is -1.65°C
<h2 />Freezing point depression
Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all less than that of the pure solvent and is directly proportional to the molality of the solute
Is melting point elevation or depression?
Boiling point elevation is that the raising of a solvent's boiling point due to the addition of a solute. Similarly, melting point depression is the lowering of a solvent's freezing point due to the addition of a solute. In fact, because the boiling point of a solvent increases, its melting point decreases
Learn more about freezing point depression :
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Answer:
i. Molar mass of glucose = 180 g/mol
ii. Amount of glucose = 0.5 mole
Explanation:
<em>The volume of the glucose solution to be prepared</em> = 500
<em>Molarity of the glucose solution to be prepared</em> = 1 M
i. Molar mass of glucose () = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol
ii.<em> mole = molarity x volume</em>. Hence;
amount (in moles) of the glucose solution to be prepared
= 1 x 500/1000 = 0.5 mole