Neutralisation of phosphorous acids
Provide solution of this question. To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3 PO3) the volume of 0.1 M aqueous KOH solution required is: (a) 40 mL (b) 20 mL (c) 10 mL (d) 60 mL
What do you mean by the term alum? Also illustrate its uses?
Choose the right answer from following. The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular weight of the substance will be: (a) 18.0 (b) 342 (c) 60 (d) 180
The important methods for the preparation of alcohol on large-scale are given below:  
Help me to go through this problem. Molarity of a solution containing 1g NaOH in 250ml of solution: (a) 0.1M (b) 1M (c) 0.01M (d) 0.001M
37% weight of HCl and density is 1.1g/ml. find molarity of HCl
Write a short note on the IUPAC name of the benzene?
The objective of this work is to separate a binary mixture and to cool down the bottom product for storage. (Check table below to see which mixture you are asked to study). 100 kmol of feed containing 10 mol percent of the lighter component enters a continuous distillation column at the m
Give me answer of this question. The concentration of an aqueous solution of 0.01M CH3OH solution is very nearly equal to which of the following : (a) 0.01%CH3OH (b) 0.1%CH3OH (c) xCH3OH= 0.01 (d) 0.99MH2O (
The following mixture of hydrocarbons is obtained as one stream in a petroleum refinery. Q : Problem on Osmotic Pressure of solution The osmotic pressure of a 5% solution of cane sugar at 150oC is (mol. wt. of cane sugar = 342)(a) 4 atm (b) 3.4 atm (c) 5.07 atm (d) 2.45 atmAnswer: (c) Π = (5 x 0.0821 x 1000 x 423)/(342 x 100) = 5.07 atm
The osmotic pressure of a 5% solution of cane sugar at 150oC is (mol. wt. of cane sugar = 342)(a) 4 atm (b) 3.4 atm (c) 5.07 atm (d) 2.45 atmAnswer: (c) Π = (5 x 0.0821 x 1000 x 423)/(342 x 100) = 5.07 atm
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