Do all the question 10.2 g = 425 g × 0 0240-10.2 m acetate (solute) in H,Omass 1
ID: 902756 • Letter: D
Question
Do all the question 10.2 g = 425 g × 0 0240-10.2 m acetate (solute) in H,Omass 1H,O C,H,O, 425 810.2 substitute the values you -The-n of Man of NGM,O 425 of solution-mass NaHO, 4 10.2 A15 of sodium acetate in 415 got 415 of water. moud propare the solution by dissolving 10.2 g of sodi nwr Check A quick way to check your work is to substi the guanity of water in the solution is - mass of solution mass or NaCHo is to substitute the values vo alues you determi k your tute: make sure that the calculated «ieten Mas of 0-mass make sure tha d value agree Exercise 125P An experiment calls for 35.0 g of hydroch mass How many grams of HCl is this? How man into the Formule for mass percentage ol so with that given in the problem acid that is 20.2% HCI that is 20.2% HCl by 0 g of hydrochloric acid that is 20 calls for 3s.ny grams of water? 12..12.80,12.5) is this? How many grams of wate See Problems 12.49, 12.50, 12.51, ant st See Prob 12.51, and 12s2 12.51, and 1Explanation / Answer
12.5) 100 gr of solution contains 20.2 gr of HCl
35 gr of HCl = 35/100)*20.2 = 7.07 gr of HCl
water = 100 - 7.07 = 92.93 gr
12.9) (mole fraction of methanol/molefraction of ethanol) = (moles of methanol/moles of ethanol)
so 0.25/0.75 = 1/3
so molality of methanol = (moles of methanol) * (1000)/wt of the ethanol
= 1*1000/3*46 [ wt of ethanol 3*mol wt)
= 7.25 m
12.10) molality of urea given = 3.42
so 3.42 moles of urea present in 1000 gr of solvent.
density of the solution = 1.045 gr/ml
wt of urea = 3.42 * mol. wt = 3.42* 60 = 205.2 gr
wt of solution = 1000+205.2 = 1205.2 gr
volume of solution = 1205.2/1.045 = 1153.5 ml
molarity , M = moles of urea*1000/Vml of solution
= 3,42*1000/1153.5 = 2.95 M
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.