Chemistry 2 Marks Questions

1. Gas B, Higher the value of K_H lower is the solubility of gas/$\text{P} = \text{K}_{H}\text{x}$.
2. Negative deviation from Raoult’s law.

1. solution:

The solution which obey Roult's law over entire range of concentration are known as ideal solution.

For ideal solution: Enthalpy of mixing of the pure components to form the solution, Δ_mix H = 0 and volume of mixing, Δ_mix V = 0

2. Molarity: 

It is defined as the number of moles of solute present in 1000 mL of the solution. Molarity is represented by M.

$\text{Molarity (M)}=\frac{\text{Number of moles of solute}}{\text{ volume of solution in mL}}\times1000$

or

$\text{M} = \frac{\text{Weight of solute}}{\text{molar mass of solute}\times \text{Volume of solution in mL}}\times1000$

1. Mole fraction is the ratio of number of moles of one component to the total number of moles in a mixture.
2. Two solutions having same osmotic pressure at a given temperature are called Isotonic Solutions.
3. Van’t Hoff factor is expressed as:

$\text{i}=\frac{\text{Normal molar mass}}{\text{abnormal molar mass}}$

4. The solution which obeys Raoult’s law under all conditions is known as an ideal solution.

In Some cases, the rate of reaction depends not only on the reactant but may also depend on the substance present as a catalyst. This is also seen in the above reaction. The rate of reaction depends on the slowest step in case of a complex reaction.

1. $\text{H}_2\text{O}_{2} +\text{I}^- \xrightarrow{\ \ \text{Slow}\ \ }\text{H}_{2}\text{O} + \text{IO}^-$

2. $\text{H}_2\text{O}_{2} +\text{IO} \xrightarrow{\ \ \text{Fast}\ \ }\text{H}_{2}\text{O} + \text{I}^-+\text{O}_2$

1. Rate law:

$\text{Rate}=-\frac{1}{2}\frac{\text{d}\big[\text{H}_2\text{O}_2\big]}{\text{dt}}=\text{k}\big[\text{H}_2\text{O}_2\big]\big[\text{I}^-\big]$

2. The overall order of reaction:

$\text{Rate}=\text{k}\big[\text{H}_2\text{O}_2\big]\big[\text{I}^-\big]$

$\text{Order}=1+1=2$

In some cases, the rate of reaction depends not only on the reactant but may also depend on the substance present as a catalyst. This is also seen in the above reaction. The rate of reaction depends on the slowest step in case of a complex reaction.

1. $\text{H}_2\text{O}_{2} +\text{I}^- \xrightarrow{\ \ \text{Slow}\ \ }\text{H}_{2}\text{O} + \text{IO}^-$

2. $\text{H}_2\text{O}_{2} +\text{IO}^- \xrightarrow{\ \ \text{Fast}\ \ }\text{H}_{2}\text{O} + \text{I}^-+\text{O}_2$

1. Rate law:

$\text{Rate}=-\frac{1}{2}\frac{\text{d}\big[\text{H}_2\text{O}_2\big]}{\text{dt}}=\text{k}\big[\text{H}_2\text{O}_2\big]\big[\text{I}^-\big]$

2. The overall order of reaction:

$\text{Rate}=\text{k}\big[\text{H}_2\text{O}_2\big]\big[\text{I}^-\big]$

$\text{Order}=1+1=2$

3. Slowest step is the rate determining step. Hence, step 1 is rate determining step.

1. $\Delta_{\text{mix}}\text{H}=0$
   
2. $\Delta_{\text{mix}}\text{V}=0$
   

1. They obey rault’s law at all ranges of concentration. I.e. Ptotal = P₁ + P₂
   
2. Both $\Delta\text{H}$ and $\Delta\text{V}$ are zero for such solutions.
   

Molar mass of water (solvent) = 18 gmol^-1

^{$\therefore\ \text{Moles of water}=\frac{1000}{18}=55.5\text{ moles.}$}

$\therefore\ \text{Mole fraction of solute}=\frac{1}{1+55.5}=0.0177$

$\text{Now},\ \frac{\text{P}^\circ-\text{P}_{\text{s}}}{\text{P}^\circ}=\text{x}_2$

$\frac{12.3-\text{P}_\text{s}}{12.3}=0.0177$

$\Rightarrow\text{P}_\text{s}=12.08\text{ KPa}$

$\therefore$ no. of moles of C₆H₆, (ⁿC₆H₆) = 30/78 = 0.385

(molar mass of C₆H₆ = 78g)

no. of moles of,

$\text{CCl}_4 (^\text{n}\text{CCl}_4)=\frac{70}{154}=0.455$

$^\text{x}\text{C}_6\text{H}_6=\frac{^\text{n}\text{C}_6\text{H}_6}{^\text{n}\text{C}_6\text{H}_6+^\text{n}\text{CCl}_4}$

$=\frac{0.385}{0.385+0.455}=\frac{0.385}{0.84}=0.458$

$^\text{x}\text{CCl}_4=1-0.458=0.542$

$300\times\frac{25}{100}+400\times\frac{40}{100}$

= 75 + 160

= 235 g

Total amount of solution = 300 + 400 = 700 g

Therefore, mass percentage (w/w) of the solute in the resulting solution, $=\frac{235}{700}\times100\%$

= 33.57%

And, mass percentage (w/w) of the solvent in the resulting solution,

= (100 - 33.57)%

= 66.43%

$\therefore$ 6.56 × 10^-2g = K_H × 1 bar .... (i)

$\therefore$ 5.00 × 10^-2g = K_H × P .... (ii)

K_H = 6.56 × 10^-2/1 bar (from i)

K_H = 5.00 × 10^-2/P bar (from ii)

$\therefore\ \frac{6.56\times10^{-2}}{1}=\frac{5.00\times10^{-2}}{\text{P}}$

$\therefore\ \text{P}=\frac{5.00}{6.56}=0.762\text{ bar.}$

$\pi_2=\text{C}_2\text{RT}\ ....(\text{ii})$

therefore on dividing the 1 by 2 we get

$\frac{\pi_1}{\pi_2}=\frac{\text{C}_1}{\text{C}_2}$

Let us calculate the concentration of the first solution with osmatic pressure 4. 98bar

mass of glucose = 36g

molar mass of gulcose = 180g/ mol

therefore number of moles of gulcose = 36/ 180

= 0.2 moles

volume of the solution = 1 L

molarity = No. of moles of glucose/ vol.of solution

molarity = 0.2/ 1 L

C₁ = 0.2 moles/ L

$\text{C}_2=\frac{\pi_2\text{C}_1}{\pi_2}$

$\text{C}_2=\frac{1.52\times0.2}{4.98}=0.61\text{ M}$

= 92/ 23 g mol^-1 = 4mol

$\text{Molality, m}=\frac{\text{No. of gm moles of solute}}{\text{wt. of solvent in kg}}$

No. of gm moles of solute = 92/ 23 = 4

Wt. of water (solvent) = 1 kg

$\text{Molality}=\frac{4}{1}=4\text{ m.}$

1. 15 ppm (by mass) means 15 parts per million (106) of the solution.

Therefore, percent by mass $=\frac{15}{10^6}\times100\%$

= 1.5 × 10^-3%

2. Molar mass of chloroform (CHCl₃) = 1 × 12 + 1 × 1 + 3 × 35.5

= 119.5 g mol^-1

Now, according to the question,

15 g of chloroform is present in 106 g of the solution.

i.e., 15 g of chloroform is present in (106 - 15) â‰ED † 106 g of water.

Therefore, Molality of the solution = $\frac{\frac{15}{119.5}\text{mol}}{10^6\times10^{-3}\text{Kg}}$

= 1.26 × 10^-4m.

p = 760 mm Hg

k_H = 4.27 × 105 mm Hg

= According to Henry's law,

P = k_Hx

$\text{x}=\frac{\text{P}}{\text{k}_\text{H}}$

$=\frac{760\text{ mm Hg}}{4.27\times10^5\text{ mm Hg}}$

= 177.99 × 10^-5

= 178 × 10⁻⁵ (approximately)

Hence, the mole fraction of methane in benzene is 178 × 10^-5