solute concentration g solute/100 ml h2o

Mass Percentage or Volume Percentage - The mass percentage of a component in a given solution is the mass of the component per 100g of the solution. For e.g., if W A is the mass of the component A, W B is the mass of the component B in a soluti..

Other Methods of Expressing Concentration of Solution - Molality is defined as the number of moles of solute dissolved per 1000 g (1 kg) of solvent. Molality is expressed as 'm'. Mole Fraction is the ratio of number of moles of one component to the total numb..

(a) 38% solution, means 38 g of HCl in 100 g of solution. Then, Mass of the solution= 100 g Molar mass of HCl = 36.5 g mol - 1 (b) The molarity of conc. HCl sample = 12.38 mol/L Molarity of HCl s..

Solubility of iodine = 1.1 x 10 - 3 mol L - 1 This means that 1.1 x 10 - 3 mol of iodine is present in 1 L of water. Molecular mass of iodine = 254 Amount of iodine dissolved per litre of water =1.1 x 10 - 3 x 254= 0.279 g = 0.0279 g Amount of iodi..

If an electrolyte contains a higher concentration of ions, which are higher in the electrochemical series than those that are lower, then these ions get discharged in preference to the lower ones. For e.g., a solution of sodium chloride in water contains two..

Volume of sulphur dioxide gas = 350 ml at STP Mass of sulphur dioxide gas = 1 g Mass of one mole of sulphur dioxide = x g/mole Volume of 1 mole of sulphur dioxide = 22400 ml at STP Mass : Volume 1 : 350 ml x : 22400 ml Mass of 1 mole of..

As per the equation, 40 + 12 + (16 x 3) 1 mole 100g 22.4 litres 100 g 24400 ml Volume of carbon dioxide formed from 100 g of CaCO 3 = 22400 ml Volume of carbon dioxide from 3.125 g of CaCO 3 = ? CaCO 3 :..

Mass of the organic compound taken = 0.244 g Mass of CO 2 formed = 0.616 g Mass of H 2 O formed = 0.108 g We know that CO 2 = C and H 2 O = 2H44 g 12 ..

Volume of 1.6 mole of water at 373 K in gaseous state Volume of 1 mol = 18 g of liquid water (density = 1 g ml - 1 ) = 18 x 1.6 x 10 - 3 L = 0.0288 L Now work done (W) = -P(V 2 - V 1 ) = -1(48.93 - 0.0288) = - 48.90 atm L = - 48.90 x..

Let the concentrations of H 2 ( g ) and I 2 ( g ) at equilibrium be 'x' mol L - 1 . Then for the reaction, The equilibrium constant, is, But, K = 54.8. So, Thus, at equilibrium, H 2 ( g ) ..