1.How much water would be needed to completely dissolve 1.79 L of the gas at a p
ID: 885784 • Letter: 1
Question
1.How much water would be needed to completely dissolve 1.79 L of the gas at a pressure of 720 torrand a temperature of 21 C? A gas has a Henry's law constant of 0.183 M/atm .
2.To what volume should you dilute 40.0 mL of a 4.45 M KI solution so that 23.5 mL of the diluted solution contains 3.15 g of KI?
3. The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 49.0 mLof ethyl acetate (C4H8O2). This mixture is stored at 25.0 C. The vapor pressure and the densities for the two pure components at 25.0 C are given in the following table. What is the vapor pressure of the stored mixture?
Compound Vapor pressure(mmHg) Density
(g/mL) acetone 230.0 0.791 ethyl acetate 95.38 0.900
Explanation / Answer
1.)
Pressure = 720 torr = 720 /760 = 0.947 atm
Henry's law constant = K = 0.183 M/atm
molarity of water = K x P
= 0.183 x 0.947
= 0.1733 M
molarity = moles / volume
0.1733 = moles / 1.79
moles = 0.1733 x 1.79 = 0.31
moles = weight / molar mass
0.31 = weight / 18
weight of water = 5.58 g --------------------------> answer
2.)
molarity = (3.15 / 166 ) x 1000 / 23.5
= 0.8075 M
M1 V1 = M2 V2
M1 = 4.45 , V1 = 40 ml , M2 = 0.807 M , V2 = ?
4.45 x 40 = 0.8075 x V2
V2 = 220.4 ml
diluted volume = 220.4ml --------------------------> answer
3)
acetone mass = density x volume = 0.791 x 70 = 55.37 g
acetone molar mass = 58 g/mol
acetone moles = 55.37 /58 = 0.955
ethyl acetate mass = 0.900 x 49 = 44.1 g
ethyl acetate molar mass =88 g /mol
ethyl acetate moles = 44.1 /88 = 0.501
total moles = 0.955 + 0.501 = 1.002
mole fraction of acetone = 0.501 / 1.002 = 0.499
mole fraction of ethyl acetate = 1 - 0.499 = 0.5001
Pacetone = mole fraction x pure state pressure
= 0.499 x 230
= 114.77 torr
P ethyl acetate = 0.501 x 95.38
= 47.78
total pressure = 114.77 + 47.78
= 162.55 torr
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