Question

1. You need 100 mL of 10% (W/V) CuSO4 (FW 160), but only CuSO4 * H2O is available.

2. A procedure requires the use of a 10% (W/V) sodium carbonate (FW 106). Prepare 200 mL using the monohydrate form.

3. Prepare 1 mmol/L Lithium standard from lithium carbonate (Li2CO3). Give the quantity in mg/L.

Answer 1

V = 100 mL, 10% w/V

if only CuSO4* H2O is available

MW of hydrate = MWCuSO4+ MW H2O = 160 + 18 = 178

Now...

0.1 * 100 mL = 10 g of CusO4 required

this implies

mass of CusO4 required --> 178/160 * 10 = 11.125 g of hydrate must be mixed

11.125 g of CuSO4 *H2O

Q2.

V = 200 mL from monohydrate

10% w

MW of NaCO3 --> 106

MW of NaCO3*H2O --> 106 +18 = 124

now..

200 ml --> 10% --> 20 g required

20 * 124/106 = 23.396 g of Monohydrate requuired (Na2CO3*H2O)

Q3.

M = 1 mmol /L

Lithium standard

2 mol of Li per = 1 mol of Li2CO3

in 1 L --> 1 mmol required

that is, 1/2 mmol of Li2CO3

add

mass of Li2CO3 = mol *MW = 0.5*73.8909 = 36.94545 mg of Li2CO3