Calculate the molalities of some commerical reagents from thefollowing data: HCl

Question

Calculate the molalities of some commerical reagents from thefollowing data: HCl has a formula weight (amu) of 36.465, Density of thesolution(g/mL) of 1.19, Weight % of 37.2, and Molarity of12.1. HC2H3O2 has a formula weightof 60.05, Density of 1.05, Weight % of 99.8, and Molarity of17.4. NH3(aq) has a formula weight of 17.03, Denisty of0.90, Weight % of 28.0, and Molarity of 14.8 Please show me how you got answers so I can understand how todo it...Thank You! Calculate the molalities of some commerical reagents from thefollowing data: HCl has a formula weight (amu) of 36.465, Density of thesolution(g/mL) of 1.19, Weight % of 37.2, and Molarity of12.1. HC2H3O2 has a formula weightof 60.05, Density of 1.05, Weight % of 99.8, and Molarity of17.4. NH3(aq) has a formula weight of 17.03, Denisty of0.90, Weight % of 28.0, and Molarity of 14.8 Please show me how you got answers so I can understand how todo it...Thank You!

Explanation / Answer

a ) 37.2 g of solute is present in 100 g of solution. Mass of solvent = 100 g - 37.2 g                            =62.8 g                            =0.0628 kg       molality         = 1000 *molarity / 1000 * d - molarity * molecular mass                             =1000 * 12.1 M / 1000 * 1.19 g / mL - 12.1 M * 36.465 g /mol                            =  12100   / 748. 7735 g                             =16.15 m b) 17.4 moles are present in 1000 mL of solution .      Mass of solution = 1000mL * 1.05 g / mL                                = 1050 g Mass ofsolute          =17.4 moles * 60.05 g / mol                                = 1044.87 g Mass ofsolvent         = 1050g - 1044.87 g                                 = 5.13 g Molality                     =17.4 moles /0.00513 kg                                 =    3391.8 m    c) 14.8 moles present in 1000 mL of solution .    Mass of solution = 1000 mL * 0.9 g /mL                                = 900 g Mass ofsolute          = 14.8 moles * 17.03 g / mol                                =  252.044 g Mass ofsolvent         = 900g - 252.044 g                                 = 647.656 g Molality                     =14.8 moles /0.647656 kg                                 =  22.84 m                                = 900 g Mass ofsolute          = 14.8 moles * 17.03 g / mol                                =  252.044 g Mass ofsolvent         = 900g - 252.044 g                                 = 647.656 g Molality                     =14.8 moles /0.647656 kg                                 =  22.84 m molality         =1000   * 14.8 / 1000 mL * 0.90 g / mL -14.8 moles * 17.03 g / mol                          = 1000 * 17.4 / 1000 * 1.05 - 17.4 * 60.05                         =

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