Question

(3) (i) How many mLs of a 0.5 M KH2PO4 stock solution are needed to make 250 mLs of a 2.5 mM K solution (1 mark)? (ii) How many g of Ca(NO3)2 are needed to make 300 mLs of a 1 M Ca(NO3)2 stock solution (1 mark)? (iii) What is the concentration of Nations in the -Mg nutrient treatment in mM and ppm as given in Table 1 (1 mark)? (iv) What is the concentration of Cl- ions in the -N treatment in mM and ppm as given in Table 1 (1 mark)? (v) Does the -N treatment used in the hydroponic demonstration allow a conclusive examination of the effects of N deficiency and why? (1 mark) ALL calculations must be shown in 3 (i-iv) above to obtain any marks. (5 marks total for question 3)

Answer 1

Part (i).

[KH2PO4] = [K⁺]

M1V1 = M2V2

i.e. 0.5 M * V1 = 2.5*10^-3 M * 250 mL

i.e. The required volume of 0.5 M KH2PO4 (V1) = 1.25 mL

Part (ii).

The required mass of Ca(NO3)2 = 1 mol/L * (300/1000) L * 164 g/mol = 49.2 g