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Question 13 Marks
A sugar syrup of weight 214.2 g contains 34.2 g of sugar $\left( C _{12} H _{22} O _{11}\right)$. Calculate
i. molal concentration, and
ii. mole fraction of sugar in the syrup
Answer

$\begin{aligned} & \text { i. } \text { Weight of sugar syrup }=214.2 g \\ & \text { Weight of sugar in syrup }=34.2 g \\ & \text { weight of water in syrup }=214.2-34.2=180.0 g \\ & \text { Moles of sugar }=\frac{34.2}{342}=0.1(\text { Molar mass }=342) \\ & \text { Molality }=\frac{0.1}{180} \times 1000=0.56 m \\ & \text { ii. } \text { Moles of sugar }=\frac{34.2}{342}=0.1 \\ & \text { Moles of water }=\frac{180}{18}=10 \\ & \text { Mole fraction of sugar }=\frac{0.1}{10+0.1} \\ & =0.0099\end{aligned}$
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Question 23 Marks
How do atomic radius vary in a period and in a group? How do you explain the variation?
Answer
Atomic radius increases down the group and decreases across the period. This is due to the continuous increase in the number of electronic shells or orbit numbers in the structure of atoms of the elements down the group.
From left to right across a period atomic radii generally decrease due to an increase in effective nuclear charge.
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Question 33 Marks
Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm . Calculate the characteristic velocity associated with the neutron.
Answer
$\begin{aligned} & \lambda=800 pm =800 \times 10^{-12} m=8 \times 10^{-10} \\ & m=1.675 \times 10^{-27} \\ & v =\frac{h}{m \lambda}=\frac{\left(6.626 \times 10^{-34} kgm ^2 s^{-1}\right)}{\left(1.675 \times 10^{-27} kg\right)\left(8 \times 10^{-10} m\right)} \\ & \frac{6.626}{1.675 \times 8} \times 10^{-34+37} ms^{-1}=\frac{6.626 \times 10^3}{1.675 \times 8} ms^{-1} \\ & =0.494 \times 10^3 ms^{-1}=494 ms^{-1}\end{aligned}$
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Question 43 Marks
PbO and $PbO _2$ react with HCl according to the following chemical equations:
$
\begin{aligned}
& 2 PbO+4 HCl \rightarrow 2 PbCl_2+2 H_2 O \\
& PbO_2+4 HCl \rightarrow PbCl_2+Cl_2+2 H_2 O
\end{aligned}
$
Why do these compounds differ in their reactivity?
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Question 53 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 mol 1 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}
$
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Question 63 Marks
1. Consider the same expansion, but this time against a constant external pressure of 1 atm.
2. What do you mean by entropy?
3. Air contains about $99 \%$ of $N _2$ and $O _2$ gases. Why they do not combine to form NO under the standard conditions? Standard Gibbs energy of formation of $NO ( g )$ is $86.7 kJ mol ^{-1}$.
Answer
1. We have $q =- w = p _{ ex }(8)=8$ litre-atm
2. 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.
3. According to the question, Standard Gibbs energy of formation of $NO ( g )$ is $86.7 kJ mol ^{-1}$.
As the standard Gibbs energy of formation is + ve , the reaction is non-spontaneous.
Hence, $N _2$ and $O _2$ do not combine to form NO.
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Question 73 Marks
Write Lewis symbols for the following atoms and ions: S and $S ^{2-} ; Al$ and $Al ^{3+} ; H$ and $H ^{-}$
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