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We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).

Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).

To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
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The dichloromethane (DCM) has less density than water and also the polarity of water is much more than DCM. So the mixture of water and dichloromethane will always be a heterogeneous mixture. In the mixture dichloromethane will be always up of the water layer. The volume of the separatory funnel which contains the mixture of DCM and water must have to be more than the total volume of the liquids thus the volume of the funnel will be more than (50+50) = 100mL.
The caution have to consider during the separation are-
1. The separatory funnel have to shake well with lid and have to settle down for some times until the two liquid separated.
2. The lid should be open very slowly as the vapor pressure of DCM is more and it will float on the water.
3. After this the stopcock should be opened and slowly the water will come out first followed by DCM.