2007 NJC Prelims Paper 1 FAQs

Explanation for Q7
p x 1.01 x 10^5 x V x 10^-6 = (m/M)RT
p x 1.01 x 10^-1 = (m/V)(RT/M)
density = m/V = 0.101pM/(8.31T) = 273pM/22400T

Explanation for Q8
An acid dissociation reaction is shown. It is a heterogeneous eqm where H2O is a solvent (i.e. in great excess), so [H2O] is NOT in Ka expression. Ka = ([H3O+][S2-])/[H2S].

NH4+ + H2O <--> NH3 + OH- and OH- reacts with H3O+ of the eqm.
CO32- also reacts with H3O+.
PbS is precipitated as a black solid, so [S2-] decreases.

Adding more water DOES result in greater dissociation of H2S. Ka has to be remain constant at the same temperature! When you add more water, more H2S must dissociate to keep Ka constant. HOWEVER, we can assume that changes in concentrations of H3O+, S2- and H2S are so small that they are negligible. So, A is the most suitable answer.

You can also say that there is NO ANSWER.

Explanation for Q10
Suppose 1 mol of e- passes through the solutions. 1/m mol of X and 1/n mol of Y are formed.
Let M be molar mass of X and 3M be molar mass of Y (since Mr(X) : Mr(Y) = 1 : 3).
Mass of X formed : Mass of Y formed = (1/m x M) : (1/n x 3M) = 1 : 2
(1/m) / (1/n) = (1/2) x (M/3M)
n/m = 3/2
Thus, n = (3/2)m

A Note for Q13
Solubility of Group II hydroxides and oxides increases down Group II.
Solubility of Group II sulphates decreases down Group II.

A Note for Q22
The intermediate is a diazo compound (azo = N, so diazo = 2 N atoms). The loss of N2 is favoured because of the strong N-N triple bond formed. Hence, adding I- results in nucleophilic substitution of the diazo compound and the formation of N2.

A Note for Q23
H2, in THIS CASE, reduces the benzene ring and the C=C double bond only. (NB. Not enough information about catalyst or conditions used)
If H2/Pt is used, ALL the functional groups shown will be reduced.

Explanation for Q26
CH3CHBrCH2CH3, when heated in NaOH(aq), gives but-1-ene or but-2-ene.
Trans-but-2-ene is most stable because the two -CH3 groups are further away from each other while the cis-isomer is less stable because the two -CH3 groups experience steric repulsion.
But-1-ene is the least stable isomer and is formed as the minor product (apply Saytzeff's rule).

Combustion is exothermic, so delta H is negative.


Explanation for Q37 (Errata)
HNO3 is NOT an electrophile because all its atoms are NOT electron deficient and have either duplet (i.e. H atom) or octet structures (N and O).
Na+ is NOT electron deficient because it has an octet structure.
Answer is D.