1
AIPMT 2003
MCQ (Single Correct Answer)
+4
-1
Formation of solution from two components can be considered as
(i) Pure solvent $$ \to $$ separated solvent molecules, $$\Delta $$H1
(ii) Pure solute $$ \to $$ separated solute molecules, $$\Delta $$H2
(iii) Separated solvent and solute molecules $$ \to $$ solution, $$\Delta $$H3
Solution so formed will be ideal if
A
$$\Delta $$Hsoln = $$\Delta $$H1 + $$\Delta $$H2 + $$\Delta $$H3
B
$$\Delta $$Hsoln = $$\Delta $$H1 + $$\Delta $$H2 $$-$$ $$\Delta $$H3
C
$$\Delta $$Hsoln = $$\Delta $$H1 $$-$$ $$\Delta $$H2 $$-$$ $$\Delta $$H3
D
$$\Delta $$Hsoln = $$\Delta $$H3 $$-$$ $$\Delta $$H1 $$-$$ $$\Delta $$H2
2
AIPMT 2003
MCQ (Single Correct Answer)
+4
-1
The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm$$-$$3, respectively. If the standard free energy difference $$\left( {\Delta {G^o}} \right)$$ is equal to 1895 J mol$$-$$1, the pressure at which graphite will be transformed into diamond at 298 K is
A
11.14 $$ \times $$ 108 Pa
B
11.14 $$ \times $$ 107 Pa
C
11.14 $$ \times $$ 106 Pa
D
11.14 $$ \times $$ 105 Pa
3
AIPMT 2003
MCQ (Single Correct Answer)
+4
-1
For the reaction,
C3H8(g) + 5O2(g) $$ \to $$ 3CO2(g) + 4H2O(l)
at constant temperature, $$\Delta $$H $$-$$ $$\Delta $$E is
A
+ RT
B
$$-$$3RT
C
+ 3RT
D
$$-$$ RT
4
AIPMT 2002
MCQ (Single Correct Answer)
+4
-1
Heat of combustion $$\Delta $$Ho for C(s), H2(g) and CH4(g) are $$-$$ 94, $$-$$ 68 and $$-$$213 kcal/mol, then $$\Delta $$Ho for C(s) + 2H2(g) $$ \to $$ CH4(g) is
A
$$-$$17 kcal
B
$$-$$111 kcal
C
$$-$$170 kcal
D
$$-$$85 kcal
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