Chemistry MCQs for NEET — Practice Questions with Answers

(a) 

$$\begin{gathered} \mathrm{P}={\mathrm{P}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}} \\ {\mathrm{P}}_{\mathrm{A}}={\mathrm{X}}_1 \mathrm{P}={\mathrm{X}}_2{\mathrm{P}}_{\mathrm{A}}^0 \\ \therefore \mathrm{P}=\frac{{\mathrm{P}}_{\mathrm{A}}^0{\mathrm{X}}_2}{{\mathrm{X}}_1} \end{gathered}$$

$$\begin{gathered} \mathrm{HA}\stackrel{ }{\rightleftharpoons }{\mathrm{H}}^{+}+{\mathrm{A}}^{-} \mathrm{C}(1+\mathrm{\alpha }) \\ \mathrm{C}(1-\mathrm{\alpha }) \mathrm{C\alpha } \mathrm{C\alpha } \\ \left[{\mathrm{H}}^{+}\right]={10}^{-2} \left[{\mathrm{A}}^{-}\right]={10}^{-2} \\ \left[\mathrm{HA}\right]=0.1-{10}^{-2}=0.09 \\ \mathrm{C}(1-\mathrm{\alpha })=0.09 \\ (1-\mathrm{\alpha })=\frac{0.09}{0.1}=0.9 \\ \mathrm{\alpha }=1-0.9=0.1 \\ \therefore \mathrm{\pi }=\mathrm{C}(1+\mathrm{\alpha })\mathrm{RT}=0.1(1.1)\mathrm{RT}=0.11\mathrm{RT} \end{gathered}$$

(c)$\frac{23.76-23.392}{23.76}=\frac{\mathrm{w}}{342\times 5.65}$

w= 30 g

(a) This depends on the number of ionic species present.

Step 1: The heat given by the hot water in order to come to the final temperature of the mixture will be same as the amount of heat gained by the ice to convert to water and raise the temperature of the water.

Let the whole amount of water present is converted to ice and its temperature is raised to some final temperature of mixture.

The amount of water present is 100.0 gm and the latent heat of fusion of ice is 333.6 J/gm

The amount of heat required by the ice to convert to water is,

$$\begin{gathered} {\mathrm{Q}}_1={\mathrm{m}}_{\mathrm{i}}{\mathrm{L}}_{\mathrm{i}} \\ =\left(100.0 \mathrm{gm}\right)\left(333.6 \mathrm{J}/\mathrm{gm}\right) \\ =33.36\times {10}^3 \mathrm{J} \end{gathered}$$

Step 2: The amount of water at 0$°$C is  100.0 gm and the specific heat capacity of water is .

 4.2 J/gm$°$C

The amount heat required by water at 0$°$C to raise its temperature to final temperature is,

$$\begin{gathered} {\mathrm{Q}}_2={\mathrm{m}}_{\mathrm{i}}{\mathrm{C}}_{\mathrm{w}}\left(∆{\mathrm{T}}_{\mathrm{ice}}\right) \\ =\left(100.0 \mathrm{gm}\right)\left(4.2 \mathrm{J}/\mathrm{g}°\mathrm{C}\right)\left({\mathrm{T}}_{\mathrm{f}}-0\right)°\mathrm{C} \\ =418\left({\mathrm{T}}_{\mathrm{f}}-0\right)\mathrm{J} \end{gathered}$$

Step 3: The amount of water at 25$°$C is 650.0 gm and the specific heat capacity of water is.

4.2 J/gm$°$C

The amount of heat given by the water at 25$°$Cis,

$$\begin{gathered} {\mathrm{Q}}_3={\mathrm{m}}_w{\mathrm{C}}_{\mathrm{w}}\left(∆{\mathrm{T}}_{water}\right) \\ =\left(650.0 \mathrm{gm}\right)\left(4.2 \mathrm{J}/\mathrm{g}°\mathrm{C}\right)\left(25-{\mathrm{T}}_{\mathrm{f}}\right) \\ =2730\left(25-{\mathrm{T}}_{\mathrm{f}}\right)\mathrm{J} \end{gathered}$$

Step 4: The gained by cold body is same as the heat lost by the hot body. Therefore,

    $$\begin{gathered} {\mathrm{Q}}_3={\mathrm{Q}}_1+{\mathrm{Q}}_2 \\ 2730\left(25-{\mathrm{T}}_{\mathrm{f}}\right)\mathrm{J}=33.36\times {10}^3\mathrm{J}+418\left({\mathrm{T}}_{\mathrm{f}}-0\right)\mathrm{J} \\ {\mathrm{T}}_{\mathrm{f}} = \frac{34890}{3148}=11.08°\mathrm{C} \end{gathered}$$

Therefore, the final temperature of mixture is 11.08$°$C.

${\mathrm{T}}_1=313.75 \mathrm{K}, {\mathrm{P}}_1=59.1 \mathrm{torr}, {\mathrm{T}}_2=353.15 \mathrm{K} {\mathrm{P}}_2= 298.7 \mathrm{torr}$

Substituting these values in the equation

$$\begin{gathered} \mathrm{In} \frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}=\frac{∆{\mathrm{H}}_{\mathrm{m},\mathrm{l}\to \mathrm{vap}.}}{\mathrm{R}}\left[\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right] \\ \mathrm{We} \mathrm{set}, \mathrm{In} \frac{298.7}{59.1}=\frac{∆{\mathrm{H}}_{\mathrm{m},\mathrm{l}\to \mathrm{vap}.}}{\mathrm{R}}\left[\frac{1}{313.75}-\frac{1}{353.15}\right] \\ ∆{\mathrm{H}}_{\mathrm{m},\mathrm{l}\to \mathrm{vap}.}=37888 \mathrm{J} {\mathrm{mol}}^{-1} \end{gathered}$$

(d) Acc to R.L. V.P. $\Rightarrow \frac{∆\mathrm{P}}{{\mathrm{P}}_0}={\mathrm{X}}_{\mathrm{B}}$

${\mathrm{X}}_{\mathrm{B}}=\frac{0.80-0.60}{0.80}=\frac{0.20}{0.80}=0.25$

(c) As the number of particles is the highest on complete dissocoation of K₂SO₄. Hence boiling point is highest for K₂SO₄. As glucose and urea is not dissociated in solution hence boiling point is lower for these two solutions.

$$\begin{gathered} \mathrm{Now}, \frac{1}{\mathrm{M}}=\frac{1}{{\mathrm{l}}_{\mathrm{s}}}\left(\frac{1}{\mathrm{m}}+\mathrm{M}\text{'}\times {10}^{-3}\right) \\ \mathrm{For} \mathrm{the} \mathrm{dilute} \mathrm{solution}, \mathrm{we} \mathrm{can} \mathrm{consider} {\mathrm{l}}_{\mathrm{s}}=1 \\ \therefore \frac{1}{\mathrm{M}}=\left(\frac{1}{\mathrm{m}}+\mathrm{M}\text{'}\times {10}^{-3}\right) \\ \mathrm{Molarity} \mathrm{of} \mathrm{solution}=0.1\mathrm{M} \\ \therefore 10-\mathrm{M}\text{'}\times {10}^{-3}=\frac{1}{\mathrm{m}} \\ \therefore \mathrm{m}=\frac{1}{10-\mathrm{M}\text{'}\times {10}^{-3}} \\ \mathrm{For} \mathrm{urea} \mathrm{M}=60 \mathrm{and} \mathrm{for} \mathrm{glucose} \mathrm{M}=180 \\ \therefore {\mathrm{m}}_{\mathrm{glu}}=\frac{1}{10-0.180} \therefore {\mathrm{m}}_{\mathrm{urea}}=\frac{1}{10-0.060} \\ \mathrm{Ans}. {\mathrm{m}}_{\mathrm{glucose}} > {\mathrm{m}}_{\mathrm{urea} }. \mathrm{Hence}, 0.1\left(\mathrm{M}\right) \mathrm{urea} \mathrm{solutions} \mathrm{have} \mathrm{lowest} \mathrm{boiling} \mathrm{point}. \end{gathered}$$

(b) Now $∆{\mathrm{T}}_{\mathrm{b}}=118.24-118.1=0.14°$

If a gm solute be dissolved in b gm solvent, then 

$$\begin{gathered} {\mathrm{M}}_2={\mathrm{K}}_{\mathrm{b}}\times \frac{\mathrm{a}\times 1000}{\mathrm{b}\times ∆{\mathrm{T}}_{\mathrm{b}}} \\ {\mathrm{M}}_2=\frac{{\mathrm{RT}}_{\mathrm{b}}^2}{\mathrm{l}\times 1000}\times \frac{\mathrm{a}\times 1000}{\mathrm{b}\times ∆{\mathrm{T}}_{\mathrm{b}}} \\ \mathrm{or} {\mathrm{M}}_2=\frac{2\times {10}^{-3}\times {\left(391.10\right)}^2}{121}\times \frac{0.4344\times 1000}{44.16\times 0.14} \\ \mathrm{or} {\mathrm{M}}_2=178 \end{gathered}$$

Hence, mol. wt. of anthracene = 178.

(C). This is purely a definition and hence (C) is correct.