Question

A solution is made by dissolving 13.5 g of glucose (C_6H_12O_6) in 0.100 kg of water. What is the mass percentage of solute in this solution? What is the molarity of this solution? Assume that the density of approximately 1.00 g/mL A solution with a density of 0.876 g/mL contains 5.0 g of toluene (C_7H_8) and 225 g of benzene (C_6H_6). Calculate the molarity of the solution. A 2.5-g sample of groundwater was found to contain 5.4 mu g of Zn^2+. What is the concentration of Zn^2+ in parts per million? ppm = mass of solute/mass of soln Times 10^6 An aqueous solution of hydrochloric acid contains 36% HCl by mass. Calculate the mole fraction of HCl in the solution. Calculate the molality of HCl in the solution. Calculate the volume of concentrated sulfuric acid of relative density 1.84 and containing 98.0% H_2SO_4 by mass that would contain 40.0 grams of pure H_2SO_4. Case of Non Electrolytes Find the molality of a glucose solution prepared by dissolving 18.0 g of glucos

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Answer 1

Example 1 a) Given mass of glucose (solute)-13.5 g Given mass of glucose (solute)- 13.5 g' x ㄨㄧ 1000 Given mass of glucose (solute) 0.0135 kg Mass of water (solvent) - 0.100 kg Mass % of solute = Mass % of glucose- Mass % of glucose-11 1.8943 %| Mass of solute Mass of solute +Mass of solution x100 0.0135kg x100 0.0135 kg +0.1000 kg The mass percentage of solute in this solution is 11.8943 %

b) Density of the solution is 1.00 g/mL Molar mass of glucose is 180.156 g/mol Given masS No. ofmoles- Molar mass 1 mol of glucose 180.156 gmol No. of moles of glucose= 13.5 No. of moles of glucose 0.0749 mol Density Mass Mass Density Mass of solution -1135 g Volume Volume- 1L Volume of solution = 1 135 Volume of solution =1.1350 L

of moles of solute No. Volume of solution (L) 0.0749 mol Molarity of solution = Molarity of solution = 1.135 L Molarity of solution = 0.06570 mol L The molarity of the solution is 0.06570 M