Exercise 8.22 - Enhanced with Feedback Part A Raes 320-324) Section 8.2 while Yo
ID: 559501 • Letter: E
Question
Exercise 8.22 - Enhanced with Feedback Part A Raes 320-324) Section 8.2 while You may want to reference completing this problem 0.50 mol of LiNOan 6.22 T. of solution Express your answer using two significant figures. Calculate the molarity of each of the following solutions. molarity= 8.0× 10-2 M Submit My Answers Give Up Correct Molarity is moles of solute per liter of solution. Dividing the number of mcles of LiNOby the volume of the sclution gives the molarity. Part B 1.9 g C2Hsoin 2.30 L of solution Express your answer using three significant figures. molarity = 1.696 Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining Not quite. Check through your calculations: you may have made a rounding error or used the wrong number of significant figures. Part C 13.12 mg KT in 102.5 mL of solution Express your answer using four significant figures. molarity = 177 ! Submit My Answers Give Up Incorrect; Try Again; 6 attempts remainingExplanation / Answer
B)
Molar mass of C2H6O = 2*MM(C) + 6*MM(H) + 1*MM(O)
= 2*12.01 + 6*1.008 + 1*16.0
= 46.068 g/mol
mass of C2H6O = 71.9 g
we have below equation to be used:
number of mol of C2H6O,
n = mass of C2H6O/molar mass of C2H6O
=(71.9 g)/(46.068 g/mol)
= 1.561 mol
volume , V = 2.30 L
we have below equation to be used:
Molarity,
M = number of mol / volume in L
= 1.561/2.3
= 0.679 M
Answer: 0.679 M
C)
Molar mass of KI = 1*MM(K) + 1*MM(I)
= 1*39.1 + 1*126.9
= 166 g/mol
mass of KI = 13.12 mg
= 0.01312 g [using conversion 1 g = 1000 mg]
we have below equation to be used:
number of mol of KI,
n = mass of KI/molar mass of KI
=(0.01312 g)/(166 g/mol)
= 7.904*10^-5 mol
volume , V = 102.5 mL
= 0.1025 L
we have below equation to be used:
Molarity,
M = number of mol / volume in L
= 7.904*10^-5/0.1025
= 7.71*10^-4 M
Answer: 7.71*10^-4 M
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