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murzikaleks [220]
2 years ago
11

Please help me please.

Chemistry
1 answer:
tangare [24]2 years ago
3 0
Sorry I am not working today
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Calculate the freezing point of a solution containing 15 grams of kcl and 1650.0 grams of water. the molal-freezing-point-depres
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Answer is: freezing point is -0,226°C.
Answer is: the molal concentration of glucose in this solution is 1,478 m.
m(KCl) = 15 g.
n(KCl) = m(KCl) ÷ M(KCl).
n(KCl) = 15 g ÷ 74,55 g/mol.
n(KCl) = 0,2 mol
m(H₂O) = 1650 g ÷ 1000 g/kg = 1,65 kg.
b = n(KCl) ÷ m(H₂O).
b = 0,2 mol ÷ 1,65 kg = 0,122 m.
Kf(water) = 1,86°C/m.
ΔT = Kf(water) · b(solution).
ΔT = 1,86°C/m · 0,122 m.
ΔT = 0,226°C.
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