(0.48 gram) x (1mole / 4.0 gram) = 0.48/4.0 = 0.12 mole
The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
Using the equation,
Δ
= i
m
where:
Δ
= change in freezing point (unknown)
i = Van't Hoff factor
= freezing point depression constant
m = molal concentration of the solution
Molality is expressed as the number of moles of the solute per kilogram of the solvent.
Molal concentration is as follows;
MM KCl = 74.55 g/mol
molal concentration =
molal concentration = 0.1219m
Now, putting in the values to the equtaion Δ
= i
m we get,
Δ
= 2 × 1.86 × 0.1219
Δ
= 0.4536°C
So, Δ
of solution is,
Δ
= 0.00°C - 0.45°C
Δ
= - 0.45°C
Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
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When a neutral hydrogen atom loses an electron, a positively-charged particle should remain.
Answer:
Explanation:
The molecular mass of C2H6 is approximately 30 or [(2 x 12) + (6 x 1)]. Therefore the molecule is about 2.5 times as heavy as the 12C atom or about the same mass as the NO atom with a molecular mass of 30 or (14+16).