Page 52 - Fpgee Question And Answer 2024 Edition

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Atomic weight 0.75
Equivalent weight =
Number of valence Equivalents of KCl = 74.5

27
Equivalent weight = = 9gm/Eq
3 Equivalents of KCl = 0.01006

1239. (c) Mole fraction is used frequently in Milliequivalents of KCl = Equivalents x 1000
experimentation involving theoretical consideration
since it gives a measure of the relative proportion of Milliequivalents of KCl = 0.01006 x 1000
moles of each constituent in a solution. It can be
calculated as follows: Milliequivalents of KCl = 10.06 mEqs

+
grams of FeSO 4 10.06 mEqs KCl = 10.06 mEqs K = 10.06 mEqs Cl -
Moles of FeSO = Molecular weight
4
1243. (b) The volume of gas can be found by the
grams of CaCl 2
Moles of CaCl = Molecular weight following formula:
2

15.1 P1V1 = P2V2
Moles of FeSO = 151 = 0.1 moles T1 T2
4

11.1
Moles of CaCl = = 0.1 moles V2 = P1 x V1 x T2
2
111 P2 x T1

Mole Fraction of FeSO = Moles of FeSO 4 = 710 x 50 x (273 +0)
4
Moles of FeSO 4 + Moles of CaCl 2
760 x (273 +25)
0.1
Mole Fraction of FeSO = = 0.5
4
0.1 + 0.1 = 42.79 cc

1240.(b) 500 cc of 5% Dextrose solution contains:
1244. (c) The calculation of molar gas constant R
500 x 5 = 25 grams of dextrose can be calculated by:
100
PV = nRT where

1241.(b) 250 mg (0.25 gm) of cefazolin powder are
diluted with water up to the 250cc mark. To find out P = 760 mm Hg = 1 atm
V = 22.4 liters
the % of drug in the final solution:
n = 1 mole
= 100 x 0.25 = 0.1% R = ? o
250 T = (273.16 + 0) = 273.16 K

o
R = 1 x 22.4 = 0.082 atm lit/mole K
1242. (c) An equivalent weight of KCl is:
1 x 273.16
Molecular weight
Equivalent weight = 1245.(c) 74.07 liters.
Number of valence

74.5 PV = nRT where:
Equivalent weight =
1
P = 740/760 atm
Weight in gram V =?
Equivalents of KCl =
Equivalent weight n = 3 moles
R = 0.08205 lit atm / mole degK

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