Chemistry, asked by sunikalvakulam80, 1 month ago

56 gm glucose (C6H1206) is dissolved in 100 gm water what is the boiling point of the solution. Kb of water is 0.53 K molal​

Answers

Answered by rojasminsahoo0
0

Explanation:

II.C.12 Hydrogen Fluoride–Tantalum Pentafluoride

HF:TaF5 is a catalyst for various hydrocarbon conversions of practical importance. In contrast to antimony pentafluoride, tantalum pentafluoride is stable in a reducing environment. The HF:TaF5 superacid system has attracted attention mainly through the studies concerning alkane alkylation and aromatic protonation. Generally, heterogeneous mixtures such as 10:1 and 30:1 HF:TaF5 have been used because of the low solubility of TaF5 in HF (0.9% at 19 °C and 0.6% at 0 °C). For this reason, acidity measurements have been limited to very dilute solutions, and an H0 value of −18.85 has been found for the 0.6% solution. Both electrochemical studies and aromatic protonation studies indicate that the HF:TaF5 system is a weaker superacid than HF:SbF5.

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