Unit – I D |

## Problems on Solutions and Colligative Properties

- 6 g of urea was dissolved in 500 g of water. calculate the percentage by mass of urea in the solution. (1.186 % by mass) problems
- 34.2 g of glucose is dissolved in 400 g of water. Calculate the percentage by mass of glucose solution. (7.48 % by mass. problems
- A solution is prepared by dissolving a certain amount of solute in 500 g of water. The percentage by mass of a solute in a solution is 2.38. Calculate mass of solute (12.19 g)
- 58 cm
^{3}of ethyl alcohol was dissolved in 400 cm^{3}of water to form 454 cm^{3}of a solution of ethyl alcohol. Calculate percentage by volume of ethyl alcohol in water. (12.78 % by volume) - 12.8 cm
^{3}of benzene is dissolved in 16.8 cm^{3}of xylene. Calculate percentage by volume of benzene. (43.24 % by volume. - 23 g of ethyl alcohol (molar mass 46 g mol
^{-1}) is dissolved in 54 g of water (molar mass 18 g mol^{-1}). Calculate the mole fraction of ethyl alcohol and water in solution. (0.1429 and 0.8571) - 4.6 cm
^{3}of methyl alcohol is dissolved in 25.2 g of water. Calculate a) percentage by mass of methyl alcohol b) mole fraction of methyl alcohol and water. Given density of methyl alcohol = 0.7952 g cm^{-3}, and C = 12, H = 1 and O = 16. (12.68, 0.0755, 0.9245) - Calculate the mole fraction of HCl in a solution of HCl containing 24.8 % of HCl by mass. Given H = 1, Cl = 35.5 (0.0347)
- A solution of NaOH (molar mass 40 g mol
^{-1}) was prepared by dissolving 1.6 g of NaOH in 500 cm^{3}of water. Calculate the molarity of NaOH solution. (0.08 mol dm^{-3}) - 11.11 g of urea (NH2CONH2) was dissolved in 100 g of water. calculate the molarity of the solution. Given N = 14, H = 1, C = 12, O = 16. (1.852 mol kg
^{-1}). - 34.2 g of sugar was dissolved in water to produce 214.2 g of sugar syrup. Calculate molality and mole fraction of sugar in the syrup. Given C = 12, H = 1 and O = 16. (0.556 mol kg
^{-1}, 0.0099) - Calculate the mole fraction of solute in its 2 molal aqueous solution. (0.0347)
- Calculate molarity and molality of the sulphuric acid solution of density 1.198 g cm
^{-3}containing 27 % by mass of sulphuric acid (molar mass 98 g mol^{-1}). (3.77 mol kg^{-1}, 3.301 mol dm^{-3}) - Calculate the mole fraction, molality and molarity of HNO3 in a solution containing 12.2 % HNO3. Given density of HNO3 as 1.038 g cm-3, H = 1, N = 14, O = 16. (0.0396, 2.205 mol kg
^{-1}, 2.01mol dm^{-3}) - Sulphuric acid is 95.8 % by mass. Calculate mole fraction and molarity of H2SO4 of density 1.91 g cm
^{-3}. Given H = 1, S = 32, O = 16. (0.80730, 17.98 mol dm^{-3}) - Commercially available concentrated hydrochloric acid is an aqueous solution containing 38% HCl gas by mass. If its density is 1.1 g cm
^{-3}, calculate the molarity of HCl solution and also calculate the mole fraction of HCl and H2O. (11.45 mol dm^{-3}, 0.232, 0.768) - An aqueous solution of NaOH is marked 10% (w/w). The density of the solution is 1.070 g cm
^{-3}. Calculate molarity, molality and mole fraction of NaOH in water. Given Na = 23, H =1 , O = 16 (2.675 mol dm^{-3}, 2.77 mol kg^{-1}, 0.0476, 0.9523) - Battery acid 4.22 M aqueous H2SO4 solution, and has density 1.21 g cm
^{-3}. What is the molality of H2SO4. Given H = 1, S = 32, O = 16 (5.298 mol kg^{-1}) - Calculate the mole fraction and molality of HNO3 in a solution containing 12.2 % HNO3. Given atomic masses H = 1, N = 14 and O = 16.
- Calculate molarity and molality of 6.3 % solution of nitric acid having density 1.04 g cm
^{-3}. Given atomic masses H = 1, N = 14 and O = 16. - A solution of glucose in water is labelled as 10 % (w/w). Calculate a) molality and b) molarity of the solution. Given the density of the solution is 1.20 g mL
^{-1}and molar mass of glucose is 180 g mol^{-1}. - Sulphuric acid is 95.8 % by mass. Calculate the mole fraction and molarity of H2SO4. Density of H2SO4 solution is 1.91 g cm
^{-3}. - The density of 5.35 M H2SO4 solution is 1.22 g cm
^{-3}. What is molality of a solution? - 10.0 g KCl is dissolved in 1000 g of water. If the density of the solution is 0.997 g cm
^{-3}, calculate a) molarity and b) molality of the solution. Atomic masses K = 39 g mol^{-1}, Cl = 35.5 g mol^{-1}.

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