80 WATER AND SOLUTION
If now we want to find the weight of materials necessary to
provide 400 grams of NaOH we translate this into F.W. From
inspection of the equation we find the number of F.W. of each
material and then translate into grams:
400 grams NaOH = 400/40 = 10 F.W.
Na 2 CO 3 required = J X 10 = 5 F.W. = 5 X 106 = 530 grams
Ca(OH) 2 required = h X 10 = 5 F.W. = 5 X 74 = 370 grams
PROBLEMS- How many times normal is each of the following solu-
tions?
(a) molal H 2 SO 4.
(6) formal A1 2 (SO 4 ) 3.
(c) molal A1C1 3.
(d) 0.1 molal Na 3 PO 4.
(e) 3 molal H 2 SO 4.
(/) 6 molal HN0 3.
(g) 2.5 formal MgSO 4. - How many times formal are the following solutions?
(a) 6iVHCl.
(b) normal Na 3 PO 4.
(c) 0.5 JV K 2 SO 4.
(d) O.liVIQCFeCsNe).
(e) 0.4iV Ba(OH) 2.
(/) normal CuSO 4. - (a) What is the normality of a solution of HC1 contain-
ing 39 per cent by weight of HC1 and of specific gravity 1.19?
(6) of a solution of HC1 containing 20 per cent by weight
and of specific gravity 1.12?
(c) of a solution of HN0 3 containing 68.6 per cent by
weight and of specific gravity 1.41?
(d) of sulphuric acid containing 96 per cent H 2 SO 4 and of
specific gravity 1.84? - What is the percentage by weight of the following
solutions?
(a) 6iV HN0 3 , specific gravity 1.195.
(fc) 6iV HC1 specific gravity 1.100.
(c) 6iV H 2 SO 4 , specific gravity 1.181.
(d) 0.5N H(C 2 H 3 O 2 ), specific gravity 1.00.