NEET Practice Questions (MCQs) with Answers & Solutions

A good lubricant generally possesses the following characteristics:

• A high boiling point and low freezing point (in order to stay liquid within a wide range of temperature)
• A high viscosity index
• Thermal stability
• Hydraulic stability
• Demulsibility
• Corrosion prevention
• A high resistance to oxidation

$$\begin{gathered} {\mathrm{d}}_{\mathrm{sol}.}=\frac{\mathrm{mass}}{\mathrm{volume}} \mathrm{N}=\frac{\mathrm{W}}{\mathrm{E}}\times \frac{1}{\mathrm{v}\left(\mathrm{L}\right)} \\ {\mathrm{V}}_{\mathrm{sol}.}=\frac{100}{0.6}\mathrm{ml} \mathrm{N}=\frac{35}{35}\times \frac{1000\times 0.6}{100}=6 \end{gathered}$$

The total vapor pressure of the mixture is less than the sum of the vapor pressures of the pure components (150 torr for A and 240 torr for 2 mol of B), which indicates a negative deviation from Raoult's law. This suggests that there are attractive intermolecular forces between A and B molecules, leading to a lower vapor pressure than expected.

$$\begin{gathered} {\mathrm{\Delta T}}_{\mathrm{b}}=i{\mathrm{K}}_{\mathrm{b}}\mathrm{m}; \\ 0.75 = i\times 5\times 0.1 \\ i=1.5; \mathrm{\alpha }=\frac{1.5-1}{2-1}=0.5 \\ {\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{\alpha }}^2\mathrm{C}}{1-\mathrm{\alpha }}=\frac{{\left(0.5\right)}^2\times 0.1}{1-0.5}=5\times {10}^{-2} \end{gathered}$$

For a non-volatile solute, the freezing point depression (ΔTf) is given by ΔTf = Kf * m, where Kf is the cryoscopic constant and m is the molality of the solution. Given that the degree of association is 50%, the van't Hoff factor (i) is 2. Substituting the values, ΔTf = 1.86 * (100.52/100) * 2 = 3.72°C. Therefore, the freezing point will be 100.52 - 3.72 = 96.8°C or -1.86°C.

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{ik}}_{\mathrm{f}}\mathrm{m} \\ 0.93=1\times 1.86\times \frac{36}{1.2\mathrm{M}}; \mathrm{M}=60 \end{gathered}$$

m is 60 gram , so 4th is right ans

According to henry's law

S=K_H X p (S=conc. of O₂ dissolved)

$$\begin{gathered} S=1.4 \times {10}^{-3}\times 0.5=7\times {10}^{-4}\mathrm{mol}/\mathrm{L}=\frac{\mathrm{w}/{\mathrm{M}}_{\mathrm{w}}}{{\mathrm{V}}_{\mathrm{L}}} \\ \Rightarrow \mathrm{w}=7\times {10}^{-4}\times {\mathrm{M}}_{\mathrm{w}}\times {\mathrm{V}}_{\mathrm{L}} \\ =7\times {10}^{-4}\times 32\times 0.1 \\ =22.4\times {10}^{-4}\mathrm{g}=2.24 \mathrm{mg} \end{gathered}$$

$$\begin{gathered} 5\mathrm{x} \to {\mathrm{x}}_5 \\ \mathrm{DOA}, \infty \frac{1-\mathrm{i}}{1-{\displaystyle \frac{1}{\mathrm{n}}}} \\ \Rightarrow 1=\frac{1-\mathrm{i}}{1-{\displaystyle \frac{1}{5}}} \\ \Rightarrow \frac{4}{5}=1-\mathrm{i} \\ \Rightarrow \mathrm{i}=1-\frac{4}{5}=\frac{1}{5}=0.2 \end{gathered}$$

$$\begin{gathered} \mathrm{\pi }=\mathrm{iCRT}=\mathrm{i}\times \frac{\mathrm{n}}{\mathrm{V}}\times \mathrm{RT} \\ \mathrm{n}=\frac{\mathrm{\pi }\times \mathrm{V}}{\mathrm{i}\times \mathrm{R}\times \mathrm{T}}=\frac{0.5\times 2}{2.47\times 0.0821\times 300}=0.0164 \mathrm{mol} \\ \therefore \mathrm{w}=0.0164\times 111 \mathrm{g} \\ =1.820 \mathrm{g} \end{gathered}$$