Question based on mole concept
Help me to solve this Question. The number of moles of SO2Cl2 in 13.5 gm is in is : (a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4
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.
Help me to solve this problem. An aqueous solution of glucose was prepared by dissolving 18 g of glucose in 90 g of water. The relative lowering in vapour pressure is: (a) 0.02 (b)1 (c) 20 (d)180
The symmetry of the molecules can be described in terms of electrons of symmetry and the corresponding symmetry operations.Clearly some molecules, like H2O and CH4, are symmetric. Now w
The O.P. (Osmotic Pressure) of equimolar solution of Urea, BaCl2 and AlCl3, will be in the order:(a) AlCl3 > BaCl2 > Urea (b) BaCl2 > AlCl3 > Urea (c) Urea > BaCl2<
Choose the right answer from following. The number of moles of KCL in 1000ml of 3 molar solution is: (a)1 (b)2 (c)3 (d)1.5
Briefly describe the difference in the Mendeleev’s table and modern periodic table?
Provide solution of this question. 60 gm of Urea (Mol. wt 60) was dissolved in 9.9 moles, of water. If the vapour pressure of pure water is P0 , the vapour pressure of solution is:(a) 0.10P0 (b) 1.10P0 (c) 0.90P0 (d) 0.99P0
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
In these reaction oxygen atom of carbonyl group is replaced by either one divalent group or two monovalent groups. Reaction by ammonia derivatives: aldehydes and ketones react with a number of ammonia derivatives such as hydroxylaminem hydrazine, semicarbazide etc. in weak acidic medium.
Give me answer of this question. "Relative lowering in vapour pressure of solution containing non-volatile solute is directly proportional to mole fraction of solute". Above statement is: (a) Henry law (b) Dulong and Petit law (c) Raoult's law (d) Le-Chatelier's pri
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