Question

how many grams of solute are in each of the following solution? B 245 ml of a 1.70 MLiCL B 45ml of a 0.70m c6h12O6 A 3.0 L of A 1.8 M Al (no3)3

CHMHERNANDEZ0819A post-lecture: Post Lecture Homework Chapter 09 Problem 9.105 How many grams of solute are in each of the following solutions? 3.0 L of a 1.8 M Al(NO3), Express your answer using two significant figures. AEC OB ? m 1150.14 Check the rounding in your final answer. No credit lost. Try again Submit Previous Answers Request Answer Part B 45 ml of a 0.70 M CHO Type here to search

Answer 1

Part A 3.0L of a 1.8M AL(NO3)₂ Solution since we know that equation of mularity Malarity = moles of solute Volume of solution in Litre 7 1.8 mal = solute moles of 3.0L moles of solute = 1.8 mal x 3.0L I moles of solute = 5.4 mall Moles of solute = mass of solute in gram molar mass of Ad (NO3)₂ 54mal = mass of solute in gram 212.996 g/mal mass of solute = 54 mol x 2 12.9969 of folute = 5 mol x 2 1 2 +996 /mol Note molar mars of A1(NO3)₂ = mass of salute = 1150.170 g 242.9467/mat

Part B 45 mL of a 0.70m C6H12O6 moles of solute = molarity & Volume of solution in Litere = 0.70 mod x 45 L 31.5 mal moles of solute = 0. 0315 mol/ maps of solute in gram = moles of solute & molar mass of С, Н, О, mass of solute = 0.0315 mal x 180.156 g mal mass of folute = 5-675 g ( Note molar mass of CH12 Of = 100.156%

Parte 245 mL of a 1.70M Lied solution. moles of solute = molarity & Volume of solution in Ligne = 1.70 mal x 24.5 L T 1000 most of solute = 0.4165 mal, mass of solute in from = moles of solute & Molar mass of Lial mape of solute = 0.4165 X 42.394 x maxg mans of solute -- 17.657 g ma Note molar mass of I Lical = 42.394 g mal