Basic Chemical Engineering Knowledge: [One Question per Day] 2010-11-29
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This post was last edited by WSL01218 on 2010-11-29 08:42. 1. Dissolve 2.76 g of glycerin in 200 g of water; the freezing point of this solution is measured to be -0.279°C. Determine the molar mass of glycerin. (The Kf of water is 1.86 K·kg·mol). 2. Which of the following compounds does not have lone pairs? A H2O B NH3 C NH4+ D H2S Note: 1) All participants will receive a reward ； 2) Please hide your replies. Do not edit after replying. 1. 92.0 g/mol 2. C
Question 1: 91.7 g/mol Question 2: C
1. The molar mass concentration of glycerin is m = 2.76 g/M * 0.2 Kg; the resulting value for M is 91.7 g/mol. There are too many formulas to write here, so only the final result is shown. 2. C
No way, haha. Check the answer
Reply 1# sailan (1) Solution: Let the molar mass of glycerin be M. ΔTfp = Kfp•m(glycerin). (273K – 272.721K) = 1.86 (K•kg•mol-1) × m(glycerin). Therefore, m(glycerin) = 0.279K / 1.86 (K•kg•mol-1) = 0.15 mol•kg-1. In 200 g of water, there is 2.76 g of glycerin; thus, in 1000 g of water, the amount of glycerin is 2.76 g × 1000 g / 200 g = 13.8 g, which equals 13.8×10-3 kg. In other words, the mass of 0.15 moles of glycerin is 13.8×10-3 kg. The molar mass M of glycerin is given by M = 13.8×10-3 kg / 0.15 mol = 92×10-3 kg•mol-1 = 92 g•mol-1. Hence, the relative molecular weight of glycerin is 92. (2) C
1. ΔTfp = Kfp•m(glycerol) – 0.279 = 1.86(K•kg•mol-1)×m(glycerol). Therefore, m(glycerol) = 0.279K/1.86K•kg•mol-1 = 0.15mol•kg-1. There are 2.76g of glycerol in 200g of water; thus, the amount of glycerol in 1000g of water is 2.76g×1000g/200g = 13.8g, which equals 13.8×10-3kg. The molar mass of glycerol, M, is 13.8×10-3kg/0.15mol = 92×10-3kg•mol-1 = 92g•mol-1. 2. C
1. Dissolve 2.76 g of glycerol in 200 g of water. The freezing point of this solution was found to be -0.279°C. Determine the molar mass of glycerol. Solution: M = Kf * 1000 * g / (*G*^T) = 92. 2. Among the following compounds, the one that does not have lone pairs is (C).
1. 91.7 g/mol 2. C
Reply 1# sailan (1)1.86*2.76\0.279\200*1000=92 (2)NH3