Page 51 - Fpgee Question And Answer 2024 Edition

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P = 790 mm Hg V = Volume in liter = 200/1000 = 0.2L
1
V = 200 ml W Weight in gram
1
o
o
T = 303 K (273 + 30 C) n = Number of moles = M = Molecular weight
1
P = 860 mm Hg o
2
V = ? T = Temperature in K = 100 + 273 = 373
2
R = Gas constant = 0.08205 lit atm/ mole degK
o
o
T = 313 K (273 + 40 C)
2
PV = nRT
P x V x T
V = 1 1 2 W
2
P x T 1 PV = RT
2
M

790 x 200 x 313 WRT
V = PV =
2
860 x 303 M

WRT 0.30 x 0.082 x 373
V = 189.78 ml M = PV = 1 x 0.2 = 46 grams/mole
2

1234. (a) The ideal gas Law equation is described as 1236. (c) The number of moles of solute in 1000 cc
follows: of solution is defined as molarity. Therefore, 1M
CaCO solution should contain 88 grams (since 1
PV = nRT where, 3
mole of CaCO = 88 grams/mole) CaCO in 1000 cc
3 3
P = Pressure (atm) of solution.

V = Volume (liter)
n = Number of moles of gas 1000 cc 1M solution requires 88 gm CaCO
3
o
T = Temperature in K 250 cc 1M solution requires ?
o
R = 0.0821 lit x atm/ mole x K 250 x 88
1000 = 22 grams CaCO
3
At Standard Temperature and Pressure (STP),
1237. (c) The equivalent weight of the molecule can
P = 1 Atm pressure be calculated by using the following formula:
o
T = 273 K
n = 1 mole Equivalent weight = Molecular weight
Number of valence
n R T
V = Equivalent weight = 106 = 53 gm/Eq
P 2

1 x 0.0821 x 273 The equivalent weight of Na 2CO 3 is numerically half
V =
1 of its molecular weight. The valence of carbonate
ion, CO 3 , is 2 and its equivalent weight is 60/2 = 30
-2
V = 22.414 liters gm/Eq. Although the valence of Na is unity, two
atoms are present in Na 2CO 3, providing a weight of
1235. (d) The ideal gas Law is described by the 2 x 23 gm = 46gm; hence its equivalent weight is
following equation: one half of this or 46/2 = 23 g/Eq. Thus the
equivalent weight of Na 2CO 3 is 30 + 23 = 53 gm.
PV = nRT where,
1238. (a) The equivalent weight of the atom can be
P = Pressure in atm = 1 calculated by using the following formula:

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