Solution of practice
Q/ 12:
Silver nitr. 0.5% 15
Make isotonic solution
How many g of pot. Nit. Should be used ?
Solution :
0.5/100=x/15
X= 0.075 g of s. n. needed
0.075 x 0.33 (E.value) = 0.024 g of NaCl represented by S.N.
0.9/100= x/ 15
X = 0.135 g of NaCl needed in Rx
0.135 g – 0.024 g = 0.111g of NaCl should be added
E. value of Pot.nit.= 0.58
0.111/ 0.58 = 0.19 g of Pot.nit. used in Rx……………
Q/16:
Hom. HBr 1% 15
Make iso. With Boric Acid
How many g of b.a. should be used?
Solution :
1/100 = x /15
X = 0.15 g of hom.HBr needed
0.15 g x 0.17 (E. value) = 0.0255 g of NaCl represented by Hom.HBr
0.9/ 100 = x/ 15
X = 0.135 g of NaCl needed in Rx
0.135 g – 0.0255 g = 0.11g of NaCl should be added
E. value of B.A = 0.52
0.11 / 0.52 = 0.21 g of B.A should be used…………………
Q/ 19 :
Oxy.HCl ½ %
B.A solution q.s.
P.W. ad 15 ml
Make iso.
How many mls. Of 5% solution of B.A . should be used ?
Solution :
0.5 /100 = x / 15
X = 0.075 g of oxy. HCl needed in Rx
(0.075 g x 0.2 (E. value))/ 0.009 = 1.66 ml of water
15 ml – 1.66 ml = 13.3 ml of 0.9% of NaCl or of 1.73% of B.A.
C1 x V1 = C2 x V2
1.73% x 13.3 ml = 5% x V2
V2 =4.62 ml of B.A should be used in Rx………………
Q/ 23 :
Tetra.HCl 0.5%
Ep.bit. 1:1000 10 ( the solution is already isotonic)
B.A. q.s.
p.w. ad 30
make iso.
How many g of b.a. should be used?
Solution :
0.5 /100 = x / 15
X = 0.15 g of tetr. HCl needed
0.15g x 0.18 (E.value) = 0.027g of NaCl represented by tetr.HCl
30 ml ( total volume) – 10 ml Ep.bit solution = 20 ml remain to calculate the isotonicity
0.9 / 100 = x /20 ml
X = 0.18 g of NaCl needed in 20 ml
0.18 g – 0.027 g = 0.153 g of NaCl should be added
0.153 / 0.52 (E. of B.A.) = 0.294 g of B.A should be used in Rx……..