Which element would form an ionic bond with Oxygen?
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:
806.3g
Explanation:
Given parameters:
Number of moles of silver nitrate = 4.85mol
Unknown:
Mass of silver chromate = ?
Solution:
2AgNO₃ + Na₂CrO₄ → Ag₂CrO₄ + 2NaNO₃
To solve this problem, we work from the known to the unknown;
- The known specie here is AgNO₃ ;
From the balanced chemical equation;
2 moles of AgNO₃ will produce 1 mole of Ag₂CrO₄
4.85 moles of AgNO₃ will produce = 2.43moles of Ag₂CrO₄
- Mass of silver chromate produced;
mass = number of moles x molar mass
Molar mass of Ag₂CrO₄
Atomic mass of Ag = 107.9g/mol
Cr = 52g/mol
O = 16g/mol
Input the parameters and solve;
Molar mass = 2(107.9) + 52 + 4(16) = 331.8g/mol
So,
Mass of Ag₂CrO₄ = 2.43 x 331.8 = 806.3g