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So in the final sample what we need is 2 moles of `CaCl_2` dissolved in water . Since e have dry salt we need to get a mass of `CaCl_2 ` that refers to 2moles of `CaCl_2` .
Molic mass of `CaCl_2` = 110.98 g/mol
Mass required for 2 moles = 110.98g/mol*2mol = 221.96g
`M = dm^(-3) = L `
So what we need is to weigh 221.96g of `CaCl_2` from a precise balance and dissolve them well in a 1L water sample. Once all the `CaCl_2` dissolved in water you will get the 2.0 M `CaCl_2` solution.
In the dry sample `CaCl_2` is 100% pure without impurities.
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