For a given system, vapor pressure can be defined as the pressure exerted by the vapor phase in equilibrium with the condensed phase.
In this case, as water sits in an open beaker it tends to evaporate thereby forming water vapor. This vapor phase then exerts a pressure over the liquid water, which is termed as the vapor pressure of water at that temperature and pressure. As water continues to evaporate, more and more molecules escape into the vapor phase which thereby increases the vapor pressure.
Ans: The vapor pressure of water increases as the water evaporates.
The amount of water that must be added to 6.0 M silver nitrate to make 500mL of 1.2 M solution is : 2000 mL
<u>Given data :</u>
Concentration of siilver nitrate ( M₁ ) = 6.0 M
volume of solution ( V₁ ) = 500 mL
Conc of solution ( M₂ )= 1.2 M
<h3>Determine the amount of water that must be added</h3>
we will apply the equation below
M₁V₁ = M₂V₂ ---- ( 1 )
where : V₂ = V₁ + water added ---- ( 2 )
V₂ ( Final volume ) = ( M₁V₁ ) / M₂
= ( 6 * 500 ) / 1.2
= 2500 mL
Back to eqaution ( 2 )
2500 mL = 500 mL + added water
therefore ; added water = 2500 - 500
= 2000 mL
Hence we can conclude that The amount of water that must be added to 6.0 M silver nitrate to make 500mL of 1.2 M solution is : 2000 mL.
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Answer:
The concentration of chloride ions in the final solution is 3 M.
Explanation:
The number of moles present in a solution can be calculated as follows:
number of moles = concentration in molarity * volume
In 100 ml of a 2 M KCl solution, there will be (0.1 l * 2mol/l) 0.2 mol Cl⁻
For every mol of CaCl₂, there are 2 moles of Cl⁻, then, the number of moles of Cl⁻ in 50 l of a 1.5 M solution will be:
number of moles of Cl⁻ = 2 * number of moles of CaCl₂
number of moles of Cl⁻ = 2 ( 50 l * 1.5 mol / l ) = 150 mol Cl⁻
The total number of moles of Cl⁻ present in the solution will be (150 mol + 0.2 mol ) 150.2 mol.
Assuming ideal behavior, the volume of the final solution will be ( 50 l + 0.1 l) 50.1 l. The molar concentration of chloride ions will be:
Concentration = number of moles of Cl⁻ / volume
Concentration = 150.2 mol / 50.1 l = 3.0 M
