Question

A `20*g` mass of `CaCl_2` was dissolved in `700*g` of water. What is the molality of the solution with respect to `"calcium chloride"`?

Answer 1

`0.25` `molal`

Explanation:

Chemical formula of Calcium chloride= `CaCl_2`
Molar mass of `CaCl_2` =`110.98` `gmol^-1`
Mass of Calcium chloride (given) = `20g`

`Molality` = `m` = `("moles of solute")/"kg of solvent"`
Now, to calculate no. of moles:
No. of moles = `("mass")/"molar mass"`

No. of moles= `("20g")/"110.98g/mol"` = `0.180` moles
`g` of solvent = `700 g`
Converting `g` `->` `kg`
`kg` of solvent = `0.7kg`
Water is solvent while `CaCl_2` is solute
Now,

`Molality` = `m` = `("moles of solute")/"kg of solvent"`
`Molality` = `("0.180 moles")/"0.7kg"`
`Molality` = `0.25` `molal`

Answer 2

`"Concentration"-=0.257*"molal"`.........

Explanation:

`"Molality"` `-=` `"Moles of solute"/"Kilograms of solvent"`

And thus..............................................

`"molality"=((20*g)/(110.98*g*mol^-1))/(700*gxx10^-3*kg*g^-1)=0.257*mol*kg`.

Note that at (relatively!) low concentrations, the calculated `"molality"` would be almost the same as solution `"molarity"`. `"Molality"` is used because the expression is fairly independent of temperature. What is the `"molal concentration"` with respect to `"chloride ion"`?