Chemistry 3 Marks Question

1. $\frac{\text{P}^{\circ}_{A}-\text{P}_{A}}{\text{P}^{\circ}_{A}}=\frac{\text{Moles of Solvent}}{\text{Moles solute + Moles of solvent}}$

$\text{P}^{\circ}_{\text{A}}=55.3\text{ mm of Hg, P}_{\text{A}}=?$

Mass of Solvent = (100 - 10)g = 90g,

No. of Moles of solvent =$\frac{\text{90g}}{\text{18g mol}^{-1}}=\text{5 mol} $

No. of Moles of solute$\frac{\text{10}}{\text{60}}=\frac{\text{1}}{\text{6}}\text{ mol}$

$\frac{\text{55.3 - P}_{A}}{\text{55.3}}=\frac{\frac{1}{6}}{\frac{1}{6}+5}=\frac{\text{1}}{\text{31}}$

$\text{P}_{A}\text{=53.52 mm Hg}.$

2. Sodium chloride dissolves in water to form two ions whereas glucose is non–electrolyte and remains in molecular form only.

i=2

$\text{M}_{2}=\frac{\text{W}_{2}\text{R}\text{T}}{\pi\text{V}}$

$\text{M}_{2}=\frac{\text{8.95}\times\text{10}^{-3}\text{g}\times\text{0.0821 L atm mol}^{-1}\text{K}^{-1}\times\text{298 K}\times760\times1000}{0.335\text{ atm}\times35\text{ L}}$

$\text{M}_2 = 14193.3 \text{g mol}^{–1} \text{ or } 1.42\times104\text{g mol}^{–1}.$

= 1.86 K kg mol^-1 x 1.20 mol kg^–1

= 2.23 K

=(273.000 – 2.23)K = 270.7 K.

= 13.97 K kg mol^-1 x 0.292 mol kg^-1 

= 4.07 K

$\pi=\text{iCRT}$

$\Rightarrow\text{i}\times\frac{\text{n}}{\text{V}}\times\text{R}\times{T}$

$\pi=3\times\frac{\text{2.5}\times10^{-2}}{\text{174g mol}^{-1}}\times\frac{\text{1}}{\text{2L}}\times{0.0821 L}\text{ atm}\text{ K}^{1}\text{mol}^{-1}\times\text{298K}$

$\pi=5.27\times10^{-3}\text{atm}$

Alternate answer

Molar mass of KCl = 39 + 35.5 = 74.5 g

Amount of KCl = 0.54 X 74.5g = 40.05g.

1. Glucose is a nonvolatile, so it decreases the vapour pressure of solution/the number of solvent molecules escaping from the surface is reduced resulting in the decrease of vapour pressure of the solvent.

2. Let the density of solution be = d g cm^-3

Volume of solution = 1L =  1000 cm³

Mass of solution = (1000 d) g

6.90 M solution means 1 L solution contains 6.90 moles of KOH.

Mass of KOH $=\text{6.90}\times\text{56}=\text{386.4 g}$

But only 30% of the solution by mass is KOH

$\frac{\text{30}}{\text{100}}\text{cm}^{3}\times\text{(1000 d)}=\text{386.4g}$

d = 1.288g cm^-3.

^{Alternate Answer}

Strength = molarity $\times$  mol mass

$=\text{6.9}\times\text{56}=\text{386.4g/L}$

30g of solute is present in 100g of solution

1g is present in  $=\frac{\text{100}}{\text{30g}}\text{g}\text{ of solution}$

$\therefore\text{ 386.4g of solute is present in}=\frac{\text{100g}\times\text{386.4g}}{\text{30 g}}=\text{1288 g of solution}$

$\therefore\text{ Density}=\frac{\text{mass}}{\text{volume}}=\frac{\text{1288g}}{\text{1000 cm}^{3}}=\text{1.288g/cm}^{3}.$

Alternate answer

$\text{M}=\frac{\text{%}\times\text{d}\times\text{10}}{\text{M}_{B}}$

$6.9=\frac{\text{30}\times\text{d}\times\text{10}}{\text{56}}$

$\text{d}=\frac{\text{6.9}\times\text{56}}{\text{30}\times\text{10}}=\text{1.288g/cm}^{3}.$

$\therefore\ \text{Mass of solvent}=100-20=80\text{g}$

1. $\text{Molality}=\frac{\text{no. of moles of Kl}}{\text{mass of solvent (Kg)}}$

$=\frac{0.120}{0.080}=1.5\text{ mol Kg}^{-1}$

2. Density of solution = 1.202 g mL^-1

^{$\text{Volume of solution}=\frac{100}{1.202}=83.2\text{ mL}\\=0.0832\text{ L}$}​​​​​​​

$\therefore\ \text{Molarity}=\frac{0.120}{0.0832}=1.44\text{ M}$

3. No. of moles of KI = 0.120

$^\text{n}\text{H}_2\text{O}=\frac{80}{18}=4.44$​​​​​​​

$\text{x}_{\text{KI}}=\frac{0.120}{0.120+4.44}$

$=\frac{0.120}{4.560}=0.0263$

$4.98=\frac{36\times\text{R}\times300}{180\times1}=60\text{R}\ \dots(\text{i})$

$1.52=\text{C}\times\text{R}\times300=300\text{CR}\ \dots(\text{ii})$

Dividing equation (ii) by (i), we get

$\frac{300\text{ CR}}{60\text{ R}}=\frac{1.52}{4.98}$

$\text{C}=0.061\text{M}$

Molar mass of sugar = 342 g mol^-1

$\text{Molality of sugar solution}=\frac{5\times1000}{342\times100}=0.146$

$\therefore\ \Delta\text{T}_\text{f}$ for sugar solution = 273.15 - 271 = 2.15°

$\Delta\text{T}_\text{f}=\text{K}_\text{f}\times\text{m}$

$\Delta\text{T}_\text{f}=\text{K}_\text{f}\times0.146\Rightarrow\text{K}_\text{f}=2.15/0.146$

Molality of glucose solution

$=\frac{5}{180}\times\frac{1000}{100}=0.278$

(Molar mass of glucose = 180 g mol^-1)

$\Delta\text{T}_\text{f}=\text{K}_\text{f}\times\text{m}=\frac{2.15}{0.146}\times0.278=4.09^\circ$

$\therefore$ Freezing point of glucose solution

= 273.15 - 4·09 = 269.06 K.

$\therefore\ \text{M}_{\text{AB}_2}=\frac{1000\times5.1\times1}{20\times2.3}=110.87\text{ g mol}^{-1}$

$\text{M}_{\text{AB}_4}=\frac{1000\times5.1\times1}{20\times1.3}=196.15\text{ g mol}^{-1}$

Let the atomic masses of A and B are 'p' and 'q' respectively.

Then molar mass of

AB₂ = p + 2q = 110.87 g mol^-1 .... (i)

And molar mass of

AB₄ = p + 4q = 196.15 g mol^-1 .... (ii)

Substracting equation (ii) from equation(i), we

get 2q = 85.28 ⇒ q = 42.64

Putting q = 42.64 in equ. (i), we get

p = 110.87 - 85.28

p = 25.59

Thus, atomic mass of A = 25.59 g mol^-1 and atomic mass of B = 42.64 g mol^-1

K_H = 1.67 × 10⁸ Pa

$\text{P}_{\text{CO}_2}=2.5\text{ atm}=2.5\times1.01325\times10^5\text{ Pa}$

$\text{P}_{\text{CO}_2}=2.533125\times10^5\text{ Pa}$

According to Henry’s law:

$\text{P}_{\text{CO}_2}=\text{K}_{\text{H}}\text{x}$

$\Rightarrow\ \text{x}=\frac{\text{P}_{\text{CO}_2}}{\text{K}_{\text{H}}}$

$=\frac{2.533125\times10^5}{1.67\times10^8}$

= 0.00152

We can write,

$\text{x}=\frac{\text{n}_{\text{CO}_2}}{\text{n}_{\text{CO}_2+\text{n}_{\text{H}_2\text{O}}}}\approx\frac{\text{n}_{\text{CO}_2}}{\text{n}_{\text{H}_2\text{O}}}$

$\text{n}_{\text{CO}_2}$ is negligible as compared to $\text{n}_{\text{H}_2\text{O}}$ [Since, ]

In 500 mL of soda water, the volume of water = 500 mL

[Neglecting the amount of soda present]

We can write:

500 mL of water = 500 g of water

$=\frac{500}{18}\text{ mol of water}$

= 27.78 mol of water

Now,

$\frac{\text{n}_{\text{CO}_2}}{\text{n}_{\text{H}_2\text{O}}}=\text{x}$

$\frac{\text{n}_{\text{CO}_2}}{27.78}=0.00152$

$\text{n}_{\text{CO}_2}=0.042\text{ mol}$

Hence, quantity of CO₂ in 500 mL of soda water = (0.042 × 44)g

= 1.848 g

Mass percentage of C₆H₆

$=\frac{\text{Mass of C}_6\text{H}_6}{\text{Mass of C}_6\text{H}_6 +\text{ Mass of CCl}_4}\times100\%$

$=\frac{22}{22+122}\times100\%$

$=15.28\%$

$=\frac{\text{Mass of CCl}_4}{\text{Total mass of the solution}}\times100\%$

Mass percentage of CCl₄

$=\frac{\text{Mass of CCl}_4}{\text{Mass of C}_6\text{H}_6 +\text{ Mass of CCl}_4}\times100\%$

$=\frac{122}{22+122}\times100\%$

$=84.72\%$

Alternatively,

Mass percentage of CCl₄ = (100 - 15.28)% = 84.72%

$\text{As}, \ \ \ \text{P}_{\text{total}}=\text{P}_{\text{A}}+\text{P}_{\text{B}}$

$=\text{x}_{\text{A}}\text{P}_{\text{A}^\circ}+\text{x}_{\text{B}}\text{P}_{\text{B}^\circ}$

(from Raoult's law)

$=\text{x}_{\text{A}}\text{P}_{\text{A}^\circ}+(1-\text{x}_{\text{A}})\text{P}_{\text{B}^\circ}$

$=\text{P}_{\text{B}^\circ}+(\text{P}_{\text{A}^\circ}-\text{P}_{\text{B}^\circ})\text{x}_{\text{A}}$

$\Rightarrow\ 600=700+(450-700)\text{x}_{\text{A}}$

$\text{or}\ \ \ \text{x}_{\text{A}}=0.40$

$\therefore\ \text{x}_{\text{B}}=1-\text{x}_{\text{A}}=1-0.40=0.60$

$\therefore\ \text{P}_{\text{A}}=\text{x}_{\text{A}}\text{P}_{\text{A}^\circ}=0.40\times450=180\text{ mm}$

$\therefore\ \text{P}_{\text{B}}=\text{x}_{\text{B}}\text{P}_{\text{B}^\circ}=0.60\times700=420\text{ mm}$

$\therefore$ Mole fraction of A in vapour phase

$=\frac{\text{P}_{\text{A}}}{\text{P}_{\text{A}}+\text{P}_{\text{B}}}=\frac{180}{180+420}=0.30$

and, Mole fraction of B in vapour phase

= 1 - 0.30 = 0.70

^{$=\frac{80}{100}\text{ P}^\circ=0.8\text{ P}^\circ$}

Let W_g of solute is present in mixture.

Moles of solute present $=\frac{\text{W}}{40}\text{ moles}$

Molar mass of octane, C₈H₁₈

= 8 x 12 + 18 = 114 gmol^-1

^{$\therefore\ \text{Moles of octane}=\frac{114}{114}=1\text{ mol}$}

^{$\text{Now},\ \frac{\text{P}^\circ-\text{P}_\text{s}}{\text{P}^\circ}=\text{x}_2=\frac{\frac{\text{W}}{40}}{\frac{\text{W}}{40}+1}$}

^{$\frac{\text{P}^\circ-0.80\text{ P}^\circ}{\text{P}^\circ}=\frac{\frac{\text{W}}{40}}{\frac{\text{W}}{40}+1}$}

^{$1-0.80=\frac{\text{W}\times40}{40(\text{W}+40)}=\frac{\text{W}}{\text{W}+40}$}

$0.20=\frac{\text{W}}{\text{W}+40}$

0.2 W + 8 = W

8 = W(l - 0.2)

8 = 0.8 W

$\therefore\ \text{W}=\frac{8}{0.8}=10g.$

= 62 g mol^-1

Number of moles of ethylene glycol $\frac{222.6\text{ g}}{62\text{ g mol}^{-1}}$

= 3.59 mol

Therefore, molality of the solution $=\frac{3.59\text{ mol}}{0.200\text{ kg}}$

= 17.95 m

Total mass of the solution = (222.6 + 200) g

= 422.6 g

Given,

Density of the solution = 1.072 g mL^-1

Therefore, Volume of the solution $=\frac{422.6\text{ g}}{1.072\text{ g mL}^{-1}}$

= 394.22 mL

= 0.3942 × 10^-3 L

Molarity of the solution$=\frac{3.59\text{ mol}}{0.39422\times10^{-3}\text{ L}}$

= 9.11 M

Molar mass of octane (C₈H₁₈)

= 8 x 12 + 18 = 114 g mol^-1

$=\frac{35.0}{114}=0.307\text{ mol}$

= 0.458 x 105.2 KPa = 48.18 kPa

Vapour pressure of octane = x₀ x P°

$\Delta \text{T}_\text{b}=\frac{2\times0.52\text{K kg mol}^{-1}\times74.5\text{g}\times1000\text{g kg}^{-1}}{74.5\text{g mol}^{-1}\times965.5\text{g}}$

$=1.077\text{K}$

On comparing it with Raoult’s law it can be seen that partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution, only the proportionality constant K_H differs from $\text{P}_\text{A}^0,$

$\text{p}_\text{A}\propto\text{x}$

$1.5=\frac{\text{Mass of solute/311}}{11}$

$\text{or}\ \ 1.5=\frac{\text{Mass of solute}}{311}$

At STP pressure (P) = 0.987bar

According to henry's law p = k_H x

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

$=\frac{0.987}{0.0035}\text{ bar}$

= 282 bar

$0.15\text{ M}=\frac{\text{x}}{122}\times\frac{1000}{250}$

$=\frac{122}{1000}\times250\times0.15$

$=\frac{4575}{1000}=4.575\text{ g}.$

 = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16

$\text{R}=8.314\text{J K}^{-1}\text{mol}^{-1},$ i.e., it is in SI units.

$=7.483\times10^{5}\text{ Nm}^{-2}$

$=7.483\text{ atm}$

$=\frac{0.714}{0.714+5.556}$

i = 2.47

Molar mass of NaHCO₃ = 1 × 23 + 1 × 1 × 12 + 3 × 16

According to the question,

HCl reacts with Na₂CO₃ and NaHCO₃ according to the following equation.

$2\text{HCl}+\text{Na}_2\text{CO}_3\rightarrow2\text{NaCl}+\text{H}_2\text{O}+\text{CO}_2\\ \text{2 mol}\ \ \ \ \ \ \ \ \text{1 mol}$

$\text{HCl}+\text{NaHCO}_3\rightarrow\text{NaCl}+\text{H}_2\text{O}+\text{CO}_2\\ \text{1 mol}\ \ \ \ \text{1 mol}$