Question

Question #7e0de

Answer 1

93.12g

`c=n/v`

Where:

`c` = concentration

`v` =volume

`n` = no. moles

So:

`n=cv`

`n= 0.5xx2.5 = 1.25`

`M_r[KCl]=(39+35.5)=74.5`

So 1 mole weighs 74.5g

So 1.25 mol weighs `1.25xx74.5 = 93.12"g"`

(Approx `A_r` values used)