Answer:
(i) 15 ppm of CHCl3 in water means that there is 15 g of CHCl3 in 106 g of water (1 million = 106)
Percentage of CHCl3
(ii)
Here we have already find the mass % is 15 x 10-4
Number of moles of solute = mass of solute / molar mass
number of moles of CHCl3 =
Molality, m of CHCl3 in drinking water sample
Answer:
We have given,
Molar mass of heptane,
C7H16 = 100 g mol–1
Molar mass of octane,
C8H18 = 114 g mol–1
Moles of heptane
Similarly, Moles of octane
Mole fraction heptane
Mole fraction of octane
Partial vapour pressure = Mole fraction x Vap.
Pressure of pure component.
Partial vapour pressure of heptane
= 0.456 x 105.2 = 47.97 kPa
Partial vapour pressure of octane
= 0.543 x 46.8 = 25.4 kPa
Total vapour pressure of solution = 73.08 kPa.
Answer:
Consider ethylene glycol as solute and water as a solvent.
Weight of solute, WB = 222.6 g
Molar mass, MB = 24 + 6 + 32 = 62
WB = 200 g = 0.200 kg.
Moles of solute,
Molaity of ethylene glycol in H2O
Total mass of solution
= 222.6 + 200 = 422.6 g
Density of solution
= 1.072 g/ ml Volume of solution
Molarity of solution,
Consider 1 litre of solution,
Osmotic pressure, π = 1.004 bar
T = 273 K = 0.083 L atm K–1 mol–1
Volume of solution = 1 litre
Weight of solute in 1 litre = 20 g
Let the molar mas be Mb
Molar mass of non-volatile solute
= 616.7 g mol–1.