Which one of the following is most soluble in water? CH_3CH_2CH_2OH CH_3CH_2CH_2

Question

Which one of the following is most soluble in water? CH_3CH_2CH_2OH CH_3CH_2CH_2OH_2OH CH_3OH CH_3CH_2OH CH_3CH_2CH_2CH_2CH_2OH The solubility of nitrogen gas at 25 degree c and 1atm is 6.5 times 10^-4 mol/L If the partial pressure of of nitrogen gas in air is 0.76 atm what is the concentration (molarity) of dissolved nitrogen? 68 52 times 10^-4 M 5.2 times 10^-4 1.1 times 10^-5 M 3.8 times 10^-4 M 49 times 10^-4 M A solution contains 28% phosphoric acid by mass. This means that 1 L of this solution contains 28 mL of phosphoric 1 mL of this solution contains 28 of of this solution contains of phosphoric acid 100 g of this solution contains 28 g of plhosphoric acid the density of this solution is 28 g/mL 1 L of this solution has a mass of 28 g Calculate the molality of a 25.4% mass aqueous solution of phosphoric acid (H 3PO_4) 25.4 m. 3.47 2.59 m 445 m The density of the solution is needed to solve the problem. A sample of rubbing alcohol contains 142.0

Explanation / Answer

23). (C) is correct. CH3OH would be more soluble in water because compounds with shorter carbon chain are more soluble in water. Due to shorter carbon chain these compounds can easily form hydrogen bonds with water.

24). (B) is correct.

Henry’s equation, P = k C

at 1 atm,

k= P/C = 1 atm / 6.8 X 10-4 mol L-1 = 1470.588 atm/mol L-1   

So the solubility of N2 at 0.76 atm:

C = (0.76 atm) / 1470.588 = 5.2 x10-4 mol L-1

25). (C) is correct because composition given in mass basis i.e. 100 g of this solution contains 28g of phosphoric acid.

26). (B) is correct.

Let 100 g of solution.

Molality = moles of solute/kg of solvent

Moles of phosphoric acid = 25.4/98 = 0.259

Amount of solvent = 100-25.4 = 74.6 g = 0.0746 Kg

Molality = 3.47 m

27). (B) is correct.

Molecular mass of isopropanol = 60.1 g/mol

Moles of isopropanol = 142/60.1 = 2.36

Moles of water = 58.0 / 18.01 = 3.22

Mole fraction of isopropanol = 2.36 /( 2.36+3.22) = 0.423

Mole fraction of water = 3.22/ ( 2.36+3.22) = 0.577

28). (B) is correct.

Moles of water = 80/ 18.01 = 4.44

Moles of glucose = 16/180.16 = 0.0888

Mole fraction of water = 4.44 / (4.44 + 0.0888) = 0.980

By Raoult's law P = P0 xw = 25.21 * 0.980 = 24.7 torr

29). (A) is correct.

Osmotic pressure

= nRT/v

=osmotic presssure in atm
n = number of moles
R = gas constant 0.0821 atm-L/mol.K
T = Temp in kelvin
v = Volume in Lit

Molecular mass of sucrose = 342.296 g/mol

Moles of sucrose n = 2.50 / 342.296 = 0.0073

T = 65+273 = 338 K

v = 205 ml = 0.205 Lit

so , = 0.989 atm

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