Suppose you want to prepare an acenitrile solution . instead of diluting 15ml of
ID: 485853 • Letter: S
Question
Suppose you want to prepare an acenitrile solution . instead of diluting 15ml of acenitrile to a volume of 100ml, you instead mix 15ml of acenitrile with 85 ml of water. the final volume is 97.3ml not 100ml. the density of pure water is 1.00g/ml, and the density of pure acetonitrile is 0.786g/ml, at 25 °c .
What is the molarity?
A.0.981M
B.1.50M
C.2.95M
D.3.54M
WHAT IS THE MOLALITY?
A.2.67molal
b.2.81 molar
c.3.10 mol
D.3.38 molar
What is the volume percent of acenitrile?
A. 12.4%vol
B.15.4%vol
c. 16.3%volume
D.17.8 %volume
What is the weight percent of acenitrile?
A. 10.1 % wt
B. 12.2 % wt
C.15.6% wt
D.17.3%wt
What is the molarity percent of acenitrile?
A. 5.73 mol%
B. 9.70 mol%
C . 13.6 mol%
D 17.4 mol%
Explanation / Answer
Given:
Total volume of solution = 97.3 mL =0.0973 L
For Aceonitrile ( solute )
V1 = 15 mL | density = 0.786 g/mL | Mass = 15 mL*0.786 g/mL = 11.79 g | Molar mass = 41 g/mol
No. of moles = Mass /molar mass = 11.79 g / 41 g/mol = 0.287 moles
For Water (solvent)
V2 = 85 mL | density = 1 g/mL | Mass = 85 mL*1.0 g/mL = 85 g = 0.085 Kg | Molar mass = 18 g/mol
No. of moles = Mass /molar mass = 85 g / 18 g/mol = 4.72 moles
1) Molality = No. of moles of solute /volume of solution in Litres = 0.287 mol / 0.0973 L = 2.95 mol/L (or M)
Answer (c) 2.95 M
2) Molality = No. of moles of solute / Mass of solvent in Kg = 0.287 / 0.085 =3.38 mol/Kg (or m)
Answer (D) 3.38 molar
3) volume percentile of acenitrile = ( 15 mL / 97.3 mL ) * 100 % = 15.4 %
Answer (B) 15.4 % vol
4) weight percentile = (11.79 / (11.79+85) ) * 100% = 12.2 %
Answer (B) 12.2 % wt
5) molarity percent = (0.287 / (0.287+4.72) ) * 100% = 5.73 %
Answer (B) 5.73 mol %
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