xTimeout alert Ast934700Bagt ch rse Home K14 of 20 110 g of 0.170 m KBr Enter yo

Question

xTimeout alert Ast934700Bagt ch rse Home K14 of 20 110 g of 0.170 m KBr Enter your answers numerically separated by commas. s. g Submit - Part C 1.70 L of a solution tat is 11.0 % KBr by mass (the density of the solution is 1.10 g/mL) Enter your answers numerically separated by commas g. L Submit Request Answer Part D a 0 140 M solution of KBr that contains juat enough KBr to preciptate 18.0 g of AgBr trom a solution containing 0.480 mol of AgNO, Enter your answers numerically separated by commas. P11

Explanation / Answer

Part B

molality = number of moles of solute/mass of solvent in kg

Molar mass of KBr = 119 gm/mol

Number of moles of KBr = Mass/molar mass = 110/119 = 0.924 moles

molality = number of moles of solute/mass of solvent in Kg

mass of solvent in Kg = 5.435 Kg

Hence the answer will be 170g, 5435g

Part C

Density = Mass/Volume

Mass of solution = Density * Volume = 1.10 g/mL * 1700 mL = 1870g

Mass of KBr = 11/100 * 1870 = 205.7 grams

Mass of water = 1870 - 205.7 = 1664.3 grams

Volume of H2O = Mass of H2O/Density = 1513 mL = 1.513 L

If the above H2O volume doesn't work, then use the value as 1.664L

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