Monday, May 25, 2009
Pharmaceutics MCQs
FPGEE Style Pharmaceutic MCQs now available. Contact me directly to order your copy today at: kirandeepp@fpgeestudyguide.com
Sunday, May 10, 2009
Sunday, March 15, 2009
Calculations for FPGEE Students
55If a radioisotope loses 95% of its activity in 110 mins, what is its half-life?
A) 85min
B) 39 min
C) 1506 min
D) 25.5 min
kt=ln{ [A0]/[A]}; kx110= ln(100/5); k = 0.027. t½= 0.693/k= 0.693/0.027= 25.5 min.
A) 85min
B) 39 min
C) 1506 min
D) 25.5 min
kt=ln{ [A0]/[A]}; kx110= ln(100/5); k = 0.027. t½= 0.693/k= 0.693/0.027= 25.5 min.
Friday, March 13, 2009
Calculations for FPGEE Students
If a radioisotope loses 95% of its activity in 110 mins, what is its half-life?
A) 85min
B) 39 min
C) 1506 min
D) 25.5 min
A) 85min
B) 39 min
C) 1506 min
D) 25.5 min
Wednesday, March 11, 2009
General Chemistry
Calculate the molar mass of a gas which has a density of 2.615g /dm3 at 298K and 101kN/m2 (R= 8.31 J K-1 mol-1, gas molar volume = 22.4 dm3 at STP).
A)29g/mol
B)24g/mol
C)64g/mol
D)32g/mol
Ans: C
M = ρRT/P = (2.615g/dm3 x 8.31 J K-1 mol-1x 298K)/101kN/m2 = 64g/mol
A)29g/mol
B)24g/mol
C)64g/mol
D)32g/mol
Ans: C
M = ρRT/P = (2.615g/dm3 x 8.31 J K-1 mol-1x 298K)/101kN/m2 = 64g/mol
Monday, March 9, 2009
General Chemistry
Calculate the molar mass of a gas which has a density of 2.615g /dm3 at 298K and 101kN/m2 (R= 8.31 J K-1 mol-1, gas molar volume = 22.4 dm3 at STP).
A)29g/mol
B)24g/mol
C)64g/mol
D)32g/mol
A)29g/mol
B)24g/mol
C)64g/mol
D)32g/mol
Saturday, March 7, 2009
Calculations for FPGEE Students
By titration you find that 15.0 cm3 of hydrochloric acid neutralise 25.0 cm3 of a 0.100 mol/dm3 solution of sodium hydroxide. What is the concentration of hydrochloric acid?
A)1.667 mol/dm3
B) 1.667 mol /cm3
C) 0.1667 mol /cm3
D) 0.1667 mol /dm3
Ans: D: 0.1667 mol/dm3
From the equation for the neutralisation reaction: HCL + NaOH = NaCl + H2O
1 mol of HCl neutralises 1 mole on NaOH.
Using the equation : Mol = Vol x Conc
we can calculate the:
Mol of base= 25cm3 x 0.1 mol/dm3 x 10-3 dm3= 2.5 mol-3.
Mol of acid = mol of base = 2.5 mol-3 = 15 x 10-3dm3 x Conc
Concentration of acid = 0.1667 mol dm-3.
A)1.667 mol/dm3
B) 1.667 mol /cm3
C) 0.1667 mol /cm3
D) 0.1667 mol /dm3
Ans: D: 0.1667 mol/dm3
From the equation for the neutralisation reaction: HCL + NaOH = NaCl + H2O
1 mol of HCl neutralises 1 mole on NaOH.
Using the equation : Mol = Vol x Conc
we can calculate the:
Mol of base= 25cm3 x 0.1 mol/dm3 x 10-3 dm3= 2.5 mol-3.
Mol of acid = mol of base = 2.5 mol-3 = 15 x 10-3dm3 x Conc
Concentration of acid = 0.1667 mol dm-3.
Thursday, March 5, 2009
Clinical Sciences
Reversible hearing loss can by caused by which one of the following drugs?
A)Erythromycin
B) Gentamicin
C)Co-amoxiclav
D)Cephaclor
Ans: A- Erythromycin causes reversible hearing loss.
Gentamicin causes auditory damage that is not reversible/
Co-amoxiclav and cephaclor to not affect the hearing.
A)Erythromycin
B) Gentamicin
C)Co-amoxiclav
D)Cephaclor
Ans: A- Erythromycin causes reversible hearing loss.
Gentamicin causes auditory damage that is not reversible/
Co-amoxiclav and cephaclor to not affect the hearing.
Sunday, March 1, 2009
Clinical Sciences
Reversible hearing loss can by caused by which one of the following drugs?
A)Erythromycin
B) Gentamicin
C)Co-amoxiclav
D)Cephaclor
A)Erythromycin
B) Gentamicin
C)Co-amoxiclav
D)Cephaclor
Friday, February 27, 2009
Antibacterials
Which of the following antibacterial drugs alters the taste in the mouth?
A) Penicillinamine
B) Azeonatram
C) Cephalexin
D) Neomycin
Ans: C: One of the side effects of Azeonatram (Azactam) is an altered taste in the mouth.
A) Penicillinamine
B) Azeonatram
C) Cephalexin
D) Neomycin
Ans: C: One of the side effects of Azeonatram (Azactam) is an altered taste in the mouth.
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