Molarity of cane sugar solution
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.
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.
The second law states that dS ≥ (dQ/T), where dS = dQ/T for a reversible process and dS > dQ/T for an irreversible process. a. Show that since dW12 = -dW21 (dWreverse = -dWforward) for a r
give atleast two application of following colligative properties
show how a mathematical definition of entropy can be obtauined from a consideration of carnot cycle?
Give me answer of this question. A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°C are 440 mmHg for pentane and 120 mmHg for hexane. The mole fraction of pentane in the vapour phase would be: (a) 0.549 (b)
Can someone please help me in getting through this problem. The mole fraction of water in 20% aqueous solution of H2O2 is: (a) 77/68 (b) 68/77 (c) 20/80 (d) 80/20
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
Describe the net charge of a non-ionized atom?
In vapor-liquid equilibrium the relative volatility αij is defined to be the ratio of the separation or K factor for species i to that for species j, that is, αij = Ki/Kj
Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to support your answer.
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
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