Question

What mass of anhydrous solute is needed to prepare 450 mL of 0.019 M NaF(aq)?

Answer 1

We would need approximately `0.36` grams of sodium fluoride.

Explanation:

Well, we have the molarity of the solution and the volume of solution, and so we use the molarity equation, which states that,

`"molarity"="moles of solute"/"liters of solution"`

So here, we get,

`0.019 \ "M"="moles of solute"/(450 \ "mL")`

Now, we change the values into appropriate units.

`0.019 \ "M"=0.019 \ "mol/L"`

`450 \ "mL"=0.45 \ "L"`

And now, we have,

`0.019 \ "mol/L"="moles of solute"/(0.45 \ "L")`

`"moles of solute"=(0.019 \ "mol")/(color(red)cancelcolor(black)"L")*0.45color(red)cancelcolor(black)"L"`

`=0.00855 \ "mol"`

Sodium fluoride `(NaF)` has a molar mass of `41.98817 \ "g/mol"`. So here, we would need,

`0.00855color(red)cancelcolor(black)"mol"*(41.98817 \ "g")/(color(red)cancelcolor(black)"mol")=0.358998853 \ "g"`

`~~0.36 \ "g"`