freezing point 0.580 mol lactose added to 1.00 kg water

freezing point 0.580 mol lactose added to 1.00 kg water

Calculate the freezing point of each of the following solutions:

a. 0.580 mole of lactose, a nonelectrolyte, added to 1.00 kg of water

b. 45.0 g of KCl, a strong electrolyte, dissolved in 1.00 kg of water

c. 1.5 moles of K3PO4, a strong electrolyte, dissolved in 1.00 kg of water

a. 비전해질인 락토스 0.580 mol을 물 1.00 kg에 녹였을 때

b. 강전해질인 KCl 45.0 g을 물 1.00 kg에 녹였을 때

c. 강전해질인 K3PO4 1.5 mol을 물 1.00 kg에 녹였을 때

a. 0.580 mole of lactose, a nonelectrolyte, added to 1.00 kg of water

몰랄농도 = 용질 mol수 / 용매 kg수

= 0.580 mol / 1.00 kg

= 0.580 mol/kg

어는점내림. Freezing-point depression

ΔTf = i × Kf × m

( i = van't Hoff 인자 )

( 참고 https://ywpop.tistory.com/1920 )

비전해질의 i = 1 이므로,

ΔTf = 1 × (1.86 ℃/m) × 0.580 m

= 1.0788

= 1.08 ℃

물의 정상어는점 = 0℃ 이므로,

용액의 어는점 = –1.08℃

답: –1.08℃

[ 관련 글 https://ywpop.blogspot.com/2024/06/freezing-point-450-g-kcl-dissolved-in.html ]

[ 관련 글 https://ywpop.blogspot.com/2024/06/freezing-point-15-mol-k3po4-dissolved.html ]

[키워드] 어는점내림 기준, 0.580 mol lactose 1.00 kg water