2 Marks Questions

Question 12 Marks

An organic compound $'A \ ' ($molecular formula $C_3H_6O)$ is resistant to oxidation but forms a compound $'B \ ' (C_3H_8O)$ on reduction.$ 'B \ '$ reacts with $HBr$ to form a bromide $'C \ '$ which on treatment with alcoholic $KOH$ forms an alkene $'D \ ' (C_3H_6).$ Deduce the structures of $A, B, C$ and $D.$

Answer

Structure of $A, B, C$ and $D$ are deduced in the following manner.

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Question 22 Marks

Define the terms Mass percentage.

Answer

Mass percentage: The mass percentage of a component in a given solution is defined as the mass of the component per 100g of the solution.
Mass $\%$ of a component $=\frac{\text { Mass of the component in the solution }}{\text { Total mass of the solution }} \times 100$

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Question 32 Marks

Calculate the temperature at which the solution containing $54 g$ of glucose, $C_6H_{12}O_6$ in $250 g$ of water will freeze. $(K_b$ for water $= 1.86 \ K \ kg mol^{-1})$

Answer

Molecular mass of glucose
$M_B=72+12+96=180 \ g mol^{-1}$
$\Delta T_f=\frac{K_f \times w_B \times 1000}{M_B \times w_A}$
$=\frac{1.86 \times 54 \times 1000}{180 \times 250}=2.23 K$
Freezing point of solution $= T _{ f }^0-\Delta T f=273-2.23=270.77 K$

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Question 42 Marks

Answer the following:
$(a) a.$ Write the mathematical relation between rate constant and half-life of a first order reaction.
$b.$ What is collision frequency?
$(b)$ If the rate equation is given below:
Rate $= k[A]^2[B]$
then what will be the unit of its rate and rate constant?

Answer

$\text {(i)} \ a.$ $t _{1 / 2}=\frac{0.693}{k}$
$\text {(ii) Unit of rate }=mol \ L^{-1} s^{-1}$
$\text { Unit of rate constant }(k)=\frac{\text { Unit of rate }}{\text { Unit of }\left[A^2\right] \times \text { Unit of }[B]}$
$=\frac{molL^{-1} s^{-1}}{\left(molL^{-1}\right)^2\left(molL^{-1}\right)}$
$=mol^{-2} L^2 s^{-1}$

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Question 52 Marks

Why are $Fe^{2+}$ compounds easily oxidises to $Fe^{3+}$ as compared to $Mn^{+2}$ compounds.

Answer

$Mn^{2+}$ compounds are less readily converted to its $+3$ oxidation state $(Mn^{3+})$ as compared to $Fe^{2+}$ because $Mn^{2+} (3d^5)$ is more stable than $Mn^{3+} (3d^4)$ due to stable half filled $d-$orbital electronic configuration. Whereas $Fe^{3+} (3d^5)$ is more stable than $Fe^{2+} (3d^6)$ due to half filled $d-$orbitals therefore $Fe^{2+}$ gets easily converted to $Fe^{3+}$

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Question 62 Marks

Write the formulae for the following coordination compounds:
i. Potassium tetrahydroxo sincate (II)
ii. Potassium trioxalatoaluminate (III)
iii. Dichloridobis cobalt III (ethane - 1, 2 diamine)

Answer

i. $K _2\left[ Zn ( OH )_4\right]$
ii. $K _3\left[ Al \left( C _2 O _4\right)_3\right]$
iii. $\left[ CoCl _2( en )_2\right]^{+}$

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