Problem based on molarity
Choose the right answer from following. The molarity of a solution of Na2CO3 having 10.6g/500ml of solution is : (a) 0.2M (b)2M (c)20M (d) 0.02M
Basic character of amines is related to their structural arrangement. Basic strength of amines depends on the relative ease of formation of the corresponding cation by accepting a proton from the acid. Greater the stability of cation is, more is basic strength of amine.Alkyl a
How much of NaOH is needed to neutralise 1500 cm3 of 0.1N HCl (given = At. wt. of Na =23): (i) 4 g (ii) 6 g (iii) 40 g (iv) 60 g
What is the correct answer. Which of the given solutions contains highest normality: (i) 8 gm of KOH/litre (ii) N phosphoric acid (iii) 6 gm of NaOH /100 ml (iv) 0.5M H2SO4
Whether HCl is a base or an acid? Briefly state your comments?
Superphosphate has the formula CaH4(PO4)2 H2O, what is the calculation to get the percentage of Phosphorus, I need to show the calculation. I know it is 30.9737622 u in weight and 2 atoms of the formula, but not sure how to work the calculation backwards.
The surface between a liquid and a vapour distinguishes these fluids. The surface tension of liquids can be looked upon as that the property which draws a liquid together and forms a liquid vapour interface, therefore, distinguishing liquids from gases.<
Briefly describe the procedure for separating the components of the gun-powder?
Benzene and toluene form nearly ideal solutions. At 20°C, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The parial vapour pressure of benzene at 20°C for a solution containing 78g of benzene and 46g of toluene in torr is: (a) 50 (b)
Give me answer of this question. Provide solution of this question. Which one of the following is the expression of Raoult's law: (a) P-P1/P = n/n+N (b) P1-P/P = N/ N+n (c)P-P2/P1= N/ N-n (d) P1-P/P2= N-n/N
171 g of cane sugar (C12H22O11) is dissolved in one litre of water. Find the molarity of the solution: (i) 2.0 M (ii) 1.0 M (iii) 0.5 M (iv) 0.25 M Choose the right answer from above.
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