Calculate the mass of water vapour present in a room of volume 400 m3 that contains air at 27 °C on a day when the relative humi
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Answer:
Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.
Explanation:
Volume of NaOH = 1.7 ml = 0.0017 L
Molarity of NaOH = 0.0811 M
Moles of NaOH = n
n = 0.0001378 mol
According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.
Then 0.0001378 mol of NaOH will neutralize:
of sulfuric acid.
Concentration of sulfuric acid in the acid rain sample: x
Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.
Answer:
Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em> by adding 1 mol of to 1 kg of water.
Explanation:
1) Moles of NaCl ,
Mass of water = m= 1 kg = 1000 g
Moles of water =
Vapor pressure of the solution =
Vapor pressure of the pure solvent that is water =
Mole fraction of solute(NaCl)=
The vapor pressure for the NaCl solution at 17.19 Torr.
2) Moles of sucrose ,
Mass of water = m = 1 kg = 1000 g
Moles of water =
Vapor pressure of the solution =
Vapor pressure of the pure solvent that is water =
Mole fraction of solute ( glucose)=
The vapor pressure for the glucose solution at 17.19 Torr.
p = p' = 17.19 Torr
Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of to 1 kg of water.