e previous 19of My Courses! Course Home Sylabus Exercise 1371: Problems-Molarity
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e previous 19of My Courses! Course Home Sylabus Exercise 1371: Problems-Molarity My Answers Give Up Complete the fellowing tabile Answer Requested Mass Mol Volume Solute Solue Solute Solution Molarity Part B KNO 2108ml 31.0 Complete the row for NaHCO Enter your answers numerically separated by a comma The Eye Scores Course Tools eText Study Are User Setings 3000 0 140 0.140 COn 603 Solute, Solde 0 5535 My Anewers Give Up Part c Complete the ow for CaH O Enter your answers numerically separated by a comma Solute, Solution mol, mL My Answers Give UpExplanation / Answer
13.71.
molarity of NaHCO3= 0.140M
Volume of NaHCO3 =300 mL = 0.3L
Molarity= number of moles/volume inL
number of moles = MxV = 0.140x0.3= 0.042 mole
number of moles of NaHCO3= 0.042 moles
molar mass of NaHCO3 = 84.0gram/mole
mass of 1 mole of NaHCO3= 84 grams
mass of 0.042 mole of NaHCO3 = 0.042 x84 = 3.528 grams
mass of NaHCO3= 3.528 grams
number of moles of NaHCO3= 0.042 moles.
b)
mass of C12H22O11 = 60.36 grams
molar mass of C12H22O11 = 342 grams
number of moles of C12H22O11 = 60.36/342= 0.176 moles
number of moles of C12H22O11 = 0.176 moles
Molrity = number of moles/volume
volume = number of moles/molarity = 0.176/0.14= 1.257 L = 1257 mL
volume of C12H22O11= 1257 mL.
13.92.
2KOH + H2SO4 ----------- K2SO4 + 2H2O
2 mole 1 mole
a) KOH H2SO4
M1= 0.255M M2=0.255M
V1 = V2= 30 mL
n1 = 2mole n2= 1 mole
for neutralisation
m1V1/n1 = M2V2/n2
0.255xV1/2 = 0.255x30/1
V1 = 60 ml
Volume of KOH= 60mL
b) KOH H2SO4
M1= 0.255M M2 = 0.130M
V1 = V2= 195ml
n1=2 n2= 1
M1V1/n1 = M2V2/n2
0.255 xV1/2= 0.130x195/1
V1= 198.8 mL
Volume of KOH= 198.8ml
c) M1V1/n1= M2V2/n2
0.255xV1/2= 0.120x45/1
V1=42,35mL
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