M solution makeup ×yDCHEM13IL Soluti × Lecture Notes-ex:ND PowerPoint Preserxe3
ID: 588650 • Letter: M
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M solution makeup ×yDCHEM13IL Soluti × Lecture Notes-ex:ND PowerPoint Preserxe3 ds/CHEM%201 3 1 L%20Solution%20Prep%20Make%20Up%20assignment%20NEVw.pdf D ACT Admission Ticket H Cel phone HealtheYo CHFHS2015 % https://fight307.wke Sp Why storms are be NEW.pdf 3/3 ADDITIONAL PROBLEMS CONT 4. Calculate the volume in milliliters of a solution required to obtain the following: a. 1.14 g of sodium chloride from a 0.0270 M solution b. 3.30 g of ethanol(C,H,OH) from a 1.50 M solution . A 25.00 mL aliquot of a 0.468 M calcium nitrate solution is mixed with 75.00 mL of 1.396 M calcium nitrate solution. Calculate the concentration of the final (combined) solution. 6. What volume of concentrated nitric acid (15.0 M) is required to make 250.0 mL of a 1.5 M nitric acid solution? 7. A 2-fold dilution of a 0.125 M HNO, solution is performed. What is the concentration of dilute solution? upExplanation / Answer
4)
a)
Molar mass of NaCl = 1*MM(Na) + 1*MM(Cl)
= 1*22.99 + 1*35.45
= 58.44 g/mol
mass of NaCl = 1.14 g
we have below equation to be used:
number of mol of NaCl,
n = mass of NaCl/molar mass of NaCl
=(1.14 g)/(58.44 g/mol)
= 1.951*10^-2 mol
we have below equation to be used:
M = number of mol / volume in L
0.027 = 1.951*10^-2/ volume in L
volume = 0.722488 L
volume = 7.225*10^2 mL
Answer: 7.225*10^2 mL
b)
Molar mass of C2H5OH = 2*MM(C) + 6*MM(H) + 1*MM(O)
= 2*12.01 + 6*1.008 + 1*16.0
= 46.068 g/mol
mass of C2H5OH = 3.30 g
we have below equation to be used:
number of mol of C2H5OH,
n = mass of C2H5OH/molar mass of C2H5OH
=(3.3 g)/(46.068 g/mol)
= 7.163*10^-2 mol
we have below equation to be used:
M = number of mol / volume in L
1.5 = 7.163*10^-2/ volume in L
volume = 0.047755 L
volume = 47.76 mL
Answer: 47.76 mL
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