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
Can someone please help me in getting through this problem. The normality of a solution of sodium hydroxide 100 ml of which includes 4 grams of NaOH is: (a) 0.1 (b) 40 (c) 1.0 (d) 0.4
When 0.01 mole of sugar is dissolved in 100g of a solvent, the depression in freezing point is 0.40o. When 0.03 mole of glucose is dissolved in 50g of the same solvent, depression in the freezing point will be:(a) 0.60o (b) 0.80o
introduction for polyhalogen compound
why o-toluidine is a weaker base than aniline?
Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal
Select the right answer of the question. Molecular weight of urea is 60. A solution of urea containing 6g urea in one litre is : (a)1 molar (b)1.5 molar (c) 0.1 molar (d) 0.01 molar
The first method begins with a well defined layer, or boundary, of solution near the center of rotation and tracks the movement of this layer to the outside of the cell as a function of time. Such a method is termed a sedimentary velocity experiment. A
The catalytic dehydrogenation of 1-butene to 1,3-butadiene, C4H8(g) = C4H6(g)+H2(g) is carried out at 900 K and 1 atm. Q : Problem on reversible process a. For a a. For a reversible process involving ideal gases in a closed system, Illustrate thatΔS = Cv ln(T2/T1) for a constant volume process ΔS = Cp ln(T2/T1) for a constant pressu
a. For a reversible process involving ideal gases in a closed system, Illustrate thatΔS = Cv ln(T2/T1) for a constant volume process ΔS = Cp ln(T2/T1) for a constant pressu
What are the benefits of soapy detergents over the soap less detergents? Briefly state the benefits?
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