Page 49 - Fpgee Question And Answer 2024 Edition
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                                                                                                +
                                                                                                                -
                                                                  38.46 mEqs NaCl = 38.46 mEqs Na  = 38.46 mEqs Cl

                                                                                M
                                                                  1223.(b)    =
                                                                             
                                                                                C p

                                                                            -3
                                                                                           6
                                                                  = 0.25 x 10 gm   =   2.5 x 10 liters
                                                                          -9
                                                                     0.1 x 10  gm
            To prepare 4 parts (3%)      2 parts (5%)
            To prepare 500 gm            ?                        1224.(d)  FP  provided  by  1%  NaCl  is  -0.58   C.  To
                                                                                                          o
                                                                  make an isotonic solution we need FP -0.52  C:
                                                                                                         o
            = 500 x 2 = 250 gm, 5% salicylic acid.
                     4                                            0.58 FP provided by   1% NaCl
                                                                  0.52 FP provided by   ?
            1221.(b)  To  solve  this  problem,  we  must  first  find
            out the amount of drug present in the final solution.   =   0.52 x 1 = 0.9% = 0.9 gm/100cc
                                                                          0.58
            The amount of atropine in 1 pint (480cc), 1 in 500
            solution:                                             1225.(d)  When  only  serum  creatinine  is  available,
                                                                  the  following  formula  (based  on  sex,  weight,  and
            = 480 x 1 = 0.96 gm of atropine.                      age  of  the  patient)  may  be  used  to  convert  this
                   500                                            value to creatinine clearance.

            Now,  this  0.96  gm  of  drug  must  be  present  in  1   CrCl =   weight(kg)x (140−age)
            teaspoonful of the drug solution, therefore we can           72 x serum creatinine (mg/dl)
            say:
                                                                         180  x (140 − 45)
                                                                          2.2
            5cc (1 teaspoonful) contains        0.96 gm           CrCl =
                                                                              72 x 1.5
            240cc solution contains             ?
                                                                  CrCl =  71.96 ml/min
            = 240 x 0.96 = 46.08 grams of atropine.
                        5                                         Therefore, for Jennifer this value should be:

            1222.(d) The amount of sodium chloride presents in    CrCl (Female) = 0.85 x male (CrCl)
            250cc of 0.9% NaCl (Normal Saline) is:
                                                                         = 0.85 x 71.96ml/minute
            = 250 x 0.9   = 2.25 grams NaCl                              = 61.17ml/minute
                    100                                                  = 61ml/min

            Total equivalents NaCl = weight in grams              1226. (c) The molecular formula for methyl chloride
                                        equivalent weight         is  CH 3Cl.  The  mass  of  a  chlorine  atom  in  methyl
                                                                  chloride is 35.5 gm/mole. The molecular weight of
            = 2.25 = 0.03846 equivalents of NaCl.                 methyl  chloride  is  50.5  gm/mole (1C  = 12,  3H = 3
               58.5                                               and 1Cl = 35.5). Therefore, the % mass of chlorine in
                                                                  methyl chloride shall be (35.5/50.5) x 100 = 70%
            mEqs of NaCl =  equivalents x 1000
                          =   0.03846 x 1000                      1227. (b)  Each  element  has  a  different  atomic
                          =   38.46 mEqs of NaCl                  number.  It  is  designated  by  the  letter  “Z.”  It
                                                                  indicates  the  number  of  protons  or  electrons  in  a
                             +
            NaCl          Na  + Cl   -                            nucleus.
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