3 Marks Question

Question 13 Marks

The density of 3 M solution of NaCl is $1.25 g mL ^{-1}$. Calculate the molality of the solution.

Answer

Given, Molarity of solution, $M =3 mol L ^{-1}$
Mass of NaCl in 1 L solution $=3 \times 58.5=175.5 g$
Mass of 1 L solution $=$ Volume $\times$ density of solution $=1000 mL \times 1.25 g / mL =1250 g$ (since density $\left.=1.25 g mL ^{-1}\right)$
Mass of water solution $=1250-175.5=1074.5 g=1.0745 kg$.
Now, Molality of solution $=\frac{\text { number of moles of solute }}{\text { mass of solvent in } kg }=\frac{3 mol}{1.0745 kg}=2.79 m$.

View full question & answer→

Question 23 Marks

Give the name and an atomic number of the inert gas atom in which the total number of d-electrons is equal to the difference in numbers of total p and s electrons.

Answer

The Kr is the first Noble gas with atomic number 36 that contains electrons in d-orbit. The electronic configuration of Kr is:
$1 s^2,2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^{10}, 4 s^2, 4 p^6$
Total number of d-electrons $=10$
Total number of p-electrons $=18$
Total number of s-electrons $=8$
$\therefore$ Difference in total number of p and s electrons $=18-8=10$
Thus, the inert gas is krypton.

View full question & answer→

Question 33 Marks

Emission transitions in the Paschen series end at orbit $n =3$ and start from orbit n and can be represeted as $v =$ $3.29 \times 10^{15}(Hz)\left[1 / 3^2-1 / n ^2\right]$ Calculate the value of n if the transition is observed at 1285 nm . Find the region of the spectrum.

Answer

$
\begin{aligned}
& v=\left(3.29 \times 10^{15} Hz\right)\left(\frac{1}{3^2}-\frac{1}{n^2}\right) \\
& \lambda=1285 nm=1285 \times 10^{-19} m=1.285 \times 10^{-16} m \\
& v=\frac{c}{\lambda}=\frac{\left(3 \times 10^8 ms^{-1}\right)}{\left(1.285 \times 10^{-6} m\right)}=2.3346 \times 10^{14} s^{-1} \\
& 2.3346 \times 10^{14}=3.29 \times 10^{15}\left[\frac{1}{3^2}-\frac{1}{n^2}\right] \\
& \frac{2.3346}{32.9}=\frac{1}{3^2}-\frac{1}{n^2} \text { or } 0.71=\frac{1}{9}-\frac{1}{n^2} \\
& \frac{1}{n^2}=\frac{1}{9}-0.071=0.111-0.071=0.04 \\
& n^2=\frac{1}{0.04}=25 \text { or } n=5
\end{aligned}
$
Paschen series lies in infrared region of the spectrum.

View full question & answer→

Question 43 Marks

Assuming the water vapor to be a perfect gas, calculate the internal energy change when 1 mol of water at $100^{\circ} C$ and 1 bar pressure is converted to the ice at $0^{\circ} C$. Given the enthalpy of fusion of ice is 6.00 kJ mol1 heat capacity of water is $4.2 J / g ^{\circ} C$.

Answer

The change take place as follows:
Step - 1: $1 mol H _2 O \left(1,100^{\circ} C \right) \longrightarrow 1 mol\left(1,0^{\circ} C \right)$ Enthalpy change $\Delta H _1$
Step - 2: $1 mol H _2 O \left(1,0^{\circ} C \right) \longrightarrow 1 mol H _2 O \left( S , 0^{\circ} C \right)$ Enthalpy change $\Delta H _2$
Total enthalpy change will be -
$
\begin{aligned}
& \Delta H=\Delta H_1+\Delta H_2 \\
& \Delta H_1=-(18 \times 4.2 \times 100) J mol^{-1} \\
& =-7560 J mol^{-1}=-7.56 k J mol^{-1} \\
& \Delta H_2=-6.00 kJ mol^{-1}
\end{aligned}
$
Therefore,
$
\begin{aligned}
& \Delta H=-7.56 kJ mol^{-1}+\left(-6.00 kJ mol^{-1}\right) \\
& =-13.56 kJ mol^{-1}
\end{aligned}
$
There is negligible change in the volume during the change form liquid to solid state.
Therefore, $p \Delta v =\Delta ng$ RT $=0$
$
\Delta H=\Delta U=-13.56 kJ mol^{-1}
$

View full question & answer→

Question 53 Marks

Consider the reactions :
a. $6 CO _2(g)+6 H _2 O ( I ) \longrightarrow C _6 H _{12} O _6( aq )+6 O _2(g)$
b. $O _3(g)+ H _2 O _2( I ) \longrightarrow H _2 O ( I )+2 O _2(g)$
Why it is more appropriate to write these reactions as :
a. $6 CO _2+12 H _2 O ( I ) \longrightarrow C _6 H _{12} O _6( aq )+6 H _2 O ( I )+6 O _2(g)$
b. $O _3(g)+ H _2 O _2( I ) \longrightarrow H _2 O ( I )+ O _2(g)+ O _2(g)$
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.

Answer

It is believed that the photosynthesis reaction occurs in two steps. In the first step, $H _2 O$ decomposes to give $H _2$ and $O _2$ in the presence of chlorophyll and the $H _2$ produced reduces $CO _2$, to $C _6 H _{12} O _6$ in the second step. During the second step, some $H _2 O$ molecules are also produced and therefore, the reaction occurs as:
a. i. $12 H _2 O ( I ) \longrightarrow 12 H _2(g)+6 O _2(g)$
ii. $6 CO _2$ (g) $+12 H _2$ (g) $\longrightarrow C _6 H _{12} O _6$ (s) $+6 H _2 O$ (I)
iii. $6 CO _2(g)+12 H _2 O ( I ) \longrightarrow C _6 H _{12} O _6(s)+6 H _2 O ( I )+6 O _2(g)$
Therefore, it is more appropriate to write the reaction for photosynthesis as (III) because it means that 12 molecules of $H _2 O$ are used per molecule of carbohydrate and $6 H _2 O$ molecules are produced per molecule of carbohydrate during the process.
b. $O _2$ is written two times in the product which suggests that 0 , is being obtained from the two reactants as:
$
\begin{aligned}
& O_3(g) \longrightarrow O_2(g)+O(g) \\
& \frac{H_2 O_2(l)+O(g) \longrightarrow H_2 O(l)+O_2(g)}{O_3(g)+H_2 O_2(l) \longrightarrow H_2 O(l)+O_2(g)+O_2(g)}
\end{aligned}
$
The path of the reaction can be studied by using $H _2 O ^{18}$ in reaction (a) or by using $H _2 O ^{18}$ or $O _3{ }^{18}$ in reaction (b).

View full question & answer→

Question 63 Marks

1. What do you mean by entropy?
2. If enthalpy of fusion and enthalpy of vaporisation of sodium metals are 2.6 and $98.2 kJ mol ^{-1}$ respectively, what is the enthalpy of sublimation of sodium?
3. Why in some reactions heat is evolved while some reactions take place only on the absorption of heat?

Answer

1. Entropy is a measure of randomness of a system. The measure of the level of disorder in a closed but changing system, a system in which energy can only be transferred in one direction from an ordered state to a disordered state. Higher the entropy, higher the disorder and lower the availability of the system's energy to do useful work.
2. According to the question, enthalpy of fusion and enthalpy of vaporisation of sodium metal is $2.6 kJ mol ^{-1}$ and $98.2 kJ^{-20}$ $mol ^{-1}$.
$\begin{aligned} & \text { We know that, enthalpy of sublimation }=\Delta_{\text {sub }} H^{\circ}=\Delta_{\text {fus }} H^{\circ}+\Delta_{\text {vap }} H^{\circ} \\ & =2.6+98.2 \\ & =100.8 kJ mol ^{-1}\end{aligned}$
3. Every substance has energy stored in it in the form of heat content.
If the heat content of reactants $\left(H_R\right)$ is greater than that of products $\left(H_P\right)$, heat is evolved.
If the heat content of reactants $\left(H_R\right)$ is less than that of products $\left(H_P\right)$, heat is absorbed.

View full question & answer→

Question 73 Marks

Predict the dipole moment of
i. a molecule of the type $AX _2$ having a linear geometry.
ii. a molecule of the type $AX _4$ having tetrahedral geometry.
iii. a molecule of the type $AX _2$ having angular geometry.
iv. a molecule of the type $AX _4$ having square planar geometry.

Answer

i. In $AX _2$ molecule with a linear geometry, the individual bond moments of $A - X$ bonds will cancel being equal in magnitude and opposite in direction. This will cause the overall dipole moment of the molecule to be 0 .
ii. In $AX _4$ molecule having tetrahedral geometry, the individual dipole moments of $A - X$ bonds will cancel out being equal in magnitude and opposite in direction. This will cause the overall dipole moment of the molecule to be zero.
iii. In $AX _2$ molecule having angular geometry, the individual bond moments of $A - X$ bonds will add up and thus the molecule will have a net non-zero dipole moment.
iv. In $AX _4$ molecule having square planar geometry the individual dipole moments of $A - X$ bonds will cancel out being equal in magnitude and opposite in direction. This will cause the overall dipole moment of the molecule to be zero.

View full question & answer→

Chemistry 3 Marks Question

Take a timed test