0
Q:

# Harmful bacteria in drinking water cannot be destroyed by

 A) Chlorination B) boiling C) adding caustic soda D) ozonisation

Explanation:

Water containing caustic soda cannot be used for drinking

Subject: Chemistry
Q:

Silver Mirror test is given by which one of the following compounds?

1. Benzophenone

2. Acetaldehyde

3. Acetone

4. Formaldehyde

 A) only 1 B) Only 2 C) 1 and 3 D) 2 and 4

Explanation:

Both Formaldehyde and Acetaldehyde will give this test

$\inline \dpi{100} \fn_jvn \small HCHO\xrightarrow[]{[Ag(NH_{3})_{2}^+]}\begin{matrix} Ag\\ Silver\: mirror \end{matrix}+Organic\: compound$

$\inline \dpi{100} \fn_jvn \small CH_{3}-CHO\xrightarrow[]{[Ag(NH_{3})_{2}^+]}\begin{matrix} Ag\\ Silver\: mirror \end{matrix}+Organic\: compound$

Filed Under: Chemistry
Exam Prep: AIEEE

0 9
Q:

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 $\inline \dpi{100} dm^{3}$ at 27°C is to a volume of 100 $\inline \dpi{100} dm^{3}$

 A) 42.3 J/ mole / K B) 38.3 J/ mole / K C) 35.8 J/ mole / K D) 32.3 J/ mole / K

Explanation:

$\inline \dpi{100} \Delta s=nRln\frac{V_{2}}{V_{1}}$

$\inline \dpi{100} \Delta s=2.303\times 2\times 8.314\log \frac{100}{10}$

$\inline \dpi{100} \Delta s=38.3\: J/mole/K$

Filed Under: Chemistry
Exam Prep: AIEEE

1 11
Q:

A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at 680 nm, the other is at :

 A) 518 nm B) 1035 nm C) 325 nm D) 743 nm

Explanation:

$\inline \dpi{100} \frac{1}{\lambda }=\frac{1}{\lambda _{1}}+\frac{1}{\lambda _{2}}$

$\inline \dpi{100} \frac{1}{355}=\frac{1}{680}+\frac{1}{\lambda _{2}}$

$\inline \dpi{100} \frac{1}{\lambda _{2}}=\frac{680-355}{355\times 680}$

$\inline \dpi{100} \Rightarrow \lambda _{2}=743\: nm$

Filed Under: Chemistry
Exam Prep: AIEEE

0 9
Q:

In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is

$\inline \dpi{100} \small 1.\: A_{2}B_{5}$

$\inline \dpi{100} \small 2.\: A_{2}B$

$\inline \dpi{100} \small 3.\: AB_{2}$

$\inline \dpi{100} \small 4.\: A_{2}B_{3}$

 A) Option 1 B) Option 2 C) Option 3 D) Option 4

Explanation:

$\inline \dpi{100} \small Z_{A}=\frac{8}{8}\; \; Z_{B}=\frac{5}{2}$

So formula of compound is $\inline \dpi{100} \small AB_{\frac{5}{2}}$

i.e., $\inline \dpi{100} \small A_{2}B_{5}$

Filed Under: Chemistry
Exam Prep: AIEEE

1 190
Q:

The degree of dissociation ($\inline \alpha$) of a weak electrolyte, $\inline A_{x}B_{y}$ is related to van't Hoff factor (i) by the expression:

$\inline 1.\: \alpha =\frac{x+y+1}{i-1}$

$\inline 2.\: \alpha =\frac{x+y-1}{i-1}$

$\inline 3.\: \alpha =\frac{i-1}{x+y+1}$

$\inline 4.\: \alpha =\frac{i-1}{x+y-1}$

 A) 1 is correct B) 2 is correct C) 3 is correct D) 4 is correct

Explanation:

$\inline \dpi{100} \fn_jvn \small Van't \: \: Hoff \: \: factor (i)=\frac{ Observed \: \: colligative\: \: property}{Normal \: \: colligative \: \: property}$

$\inline \dpi{100} \fn_jvn \small \begin{matrix} A_{x}B_{y}\\ 1-\alpha \end{matrix}\rightarrow \begin{matrix} xA^{+y}\\ x\alpha \end{matrix}+\begin{matrix} yB^{-x}\\ y\alpha \end{matrix}$

Total Moles = $\inline \dpi{100} \fn_jvn \small 1-\alpha +x\alpha +y\alpha$

$\inline \dpi{100} \fn_jvn \small =1-\alpha(x +y-1)$

$\inline \dpi{100} \fn_jvn \small i=\frac{1+\alpha(x +y-1)}{1}$

$\inline \dpi{100} \fn_jvn \small \Rightarrow \alpha =\frac{i-1}{x+y-1}$