Answer to Question #155391 in General Chemistry for Snowy Morudu

Question

A stock solution of 1000 ppm (w/V) Cu(II) was prepared from Cu(NO₃)₂.  

1. Calculate the mass of the solid needed to prepare 1.00 L.    

           

2.    Determine the volumes needed to prepare 50.0 mL calibration standards with concentrations

           of 5.00 ppm; 10.0 ppm and 25.0 ppm, respectively.     

       

Answer 1

1. 1000 ppm means that there is 1g of Cu(II) in 1000 ml.

w(Cu)=Ar(Cu)/Mr(Cu(NO3)2)

Mr(Cu(NO3)2)=64+2×14+6×16=188(g/mol)

w(Cu)=64/188=0.34

m(Cu(NO3)2)=m(Cu)/0.34= 1/0.34=2.94(g)

2. C1V1=C2V2

V1=C2V2/C1

1) V1=5×50/1000=0.25(ml)

2)V1=10×50/1000=0.5(ml)

3)V1=25×50/1000=1.25(ml)