যদি 3g গ্লুকোজ (আণবিক ভর 180) 15^(@)C তাপমাত্রায় 60g জলে দ্রবীভূত হয়, তাহলে অসমোটিক...

If 3g of glucose (molecular mass 180) is dissolved in 60g of water at 15^(@)C , then the osmotic pressure of this solution will be :

If 23 g of glucose ( molecular mass 180)is dissolved in 60ml of water at 15^(@)C , then the osmotic pressure of this solution will be

The solubility of common salt is 36.0 g in 100 g of water at 20^(@) C . If systems I , II and III contains 20.0 , 18.0 "and" 15.0 of the salt added to 50.0 g of water in each case , the vapour pressures would be in the order :

0.6 g of a gas at 15^(@)C and 745 mm Hg occupies 200cm^(3) . It occupies 182.6cm^(3) in dry state at N.T.P. Calculate the aqueous tension at 15^(@)C.

The addition of 3g of a substance to 100g C CI_(4)(M = 154 g mol^(-1)) raise the boiling point of C CI_(4) by 0.60^(@)C of K_(b)(C CI_(4)) is 5.03 kg mol^(-1) K . Calculate: (a) the freezing point depression (b) the relative lowering of vapour pressure (c) the osmotic pressure at 298K the molar mass of the substance Given K_(f)(C CI_(4)) = 31.8 kg mol^(-1) K and rho (density) of solution = 1.64g//cm^(3)

The osmotic pressure of a solution (density is 1 g mL^(-1)) containing 3 g of glucose (molecular weight =180) in 60 g of water at 15^(@)C is

100 g of water at 60^(@)C is added to 180 g of water at 95^(@)C . The resultant temperature of the mixutre is _____.