You make 1.000 L of aqueous solution that contains 35.0 g of sucrose. How many liters of water would you have to add to this solution to reduce the molarity you calculated in part A (.194 moles) by a factor of two?

1 liter

Explanation:

Molality is calculated by dividing the moles of the solute with kilograms of solvent(unit: moles / L). The molecular weight for sucrose is 343.2 g/mol and the density of water is 1kg/L. Then the molality of the solution right now is:  

molality = moles of solute / kg of solvent

molality = 35g / (342.3g/mol)   /    1L*1kg/L= 0.102 moles / 1 kg= 0.102 m

You want to make the molality into half (0.056 m) with the same number of solute(0.102 moles) but adding more solvent. Then the number of water you need to add is

molality = moles of solute / kg of solvent

0.056 m= 0.102 moles/ solvent

solvent= (0.056 moles/ kg)/ 0.102 moles= 2kg

Two kilograms of water in liters will be: 2kg / (1kg/L)= 2L

The solution already has 1L of water. Liters of water you need to add will be: 2L - 1L = 1L

< > according to newton's third law of motion, forces always act in equal but opposite pairs. another way of saying this is for every action, there is an equal but opposite reaction.< >

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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