1. If 3.01g of the vapor of a volatile liquid is able to fill a 352-mL flask at
ID: 911321 • Letter: 1
Question
1.
If 3.01g of the vapor of a volatile liquid is able to fill a 352-mL flask at 210C and 728mm Hg, what is the molecular weight of the liquid?
**Write answer in decimal form to one decimal place.**
2.
It was important that the flask be completely dry before the unknown liquid was added so that water present would not vaporize when the flask was heated. A typical single drop of liquid water has a volume of approximately 0.05mL.
Assuming the density of water is 1.0g/mL, if 6 drops of water were left in the flask how many moles of water would this be?
** Write answer in decimal form to four decimal places.**
Explanation / Answer
1) Given that
mass of vapor of volatile liquid, m = 3.01 g
volume of flask ,V = 352 mL = 0.352 L
pressure , P = 728 mm Hg = 728/760 atm = 0.958 atm ( 1 atm = 760 mm Hg)
temperature, T= 210oC = 210+273 K = 483 K
Molecular weight (or) molar mass of liquid, M = ?
Ideal gas equation is PV = nRT where R = universal gas constant = 0.0821 L.atm/K/mol
PV = (m/M) RT [ moles n = mass (m) / molarmass (M) ]
M = mRT/PV
= [(3.01g) (0.0821 L.atm/K/mol) (483 K) ]/ [(0.958) (0.352 L)]
= 353.95 g/mol
M = 353.95 g/mol
Therefore, molecular weight of the liquid = 353.95 g/mol
b) Given that
volume of single drop of liquid water = 0.05 mL
density of water = 1.0 g/mL
Hence, mass of single drop water = volume x density = 0.05 mL x 1.0 g/mL = 0.05 g
Therefore, mass of 6 drops of water = 6 x 0.05 g = 0.3 g
We know that molar mass of water = 18 g/mol
Hence, moles of 6 drops of water = mass of 6 drops of water / molar mass of water
= (0.3 g)/ (18 g/mol)
= 0.0167 moles
moles of 6 drops of water = 0.0167
Therefore, 6 drops of water are equal to 0.0167 moles of water.
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