Question

1. You need 50 g of CuSO4 * H2O (FW 178), but only the anhydrous form is available. how much of the anhydrous form is equivalent to the 50g of monohydrate?

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

3. Convert 2M NaCl (FW 58) to % (W/V).

4. What is the molarity of a 6.5% W/V NaOH (FW 40)?

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

Answer 1

1)

  

Molar mass of CuSO4.H2O,

MM = 1*MM(Cu) + 1*MM(S) + 5*MM(O) + 2*MM(H)

= 1*63.55 + 1*32.07 + 5*16.0 + 2*1.008

= 177.636 g/mol

mass(CuSO4.H2O)= 50 g

number of mol of CuSO4.H2O,

n = mass of CuSO4.H2O/molar mass of CuSO4.H2O

=(50.0 g)/(177.636 g/mol)

= 0.2815 mol

So, anhydrous form will also be 0.2815 mol

Lets find the mass of it

Molar mass of CuSO4,

MM = 1*MM(Cu) + 1*MM(S) + 4*MM(O)

= 1*63.55 + 1*32.07 + 4*16.0

= 159.62 g/mol

mass of CuSO4,

m = number of mol * molar mass

= 0.2815 mol * 159.62 g/mol

= 44.9 g

Answer: 44.9 g

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