How many grams of cacl2 would be dissolved in 1.0L of a 0.100M solution of cacl2

Solution,

molarity is the no of moles of solute per unit volume.

We can calculate amount of CaCl2 required to prepare 0.1 M CaCl2 1000 ml solution.

we know that to prepare one ltr of 1 M solution of CaCl2 111 g required

Now consider x gram will require to prepare to

so that comparing above both condition

1000ml ×1M×X g=1000ml×0.1M×111g

X= 11.1 gram

X= 11.1 g of CaCl2

Hence 11.1 g of CaCl2 would be dissolved in 1.0L of a 0.100 M solution of CaCl2

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explanation:

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