Solutions

Q: Vapour pressure of chloroform (CHCl_{3}) and dichloromethane (CH_{2}Cl_{2}) at 25^oC are 200 mm Hg and 415 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of CHCl_{3} and 40 g of CH_{2}Cl_{2} at the same temperature will be:[Molecular mass of CHCl_{3} = 119.5\ g/mol and molecular mass of CH_{2}Cl_{2} = 85 \ g/mol ]

Molar mass of CH_2Cl_2 = 12 \times 1 + 1 \times 2 + 35.5 \times 2 = 85 g mol ^{–1} Molar mass of CHCl_3 = 12 \times 1 + 1 \times 1 + 35.5 \times 3 = 119.5 g mol ^{-1} Moles of CH_2Cl_2 = 40 g /85 g mol ^{-1} = 0.47 molMoles of CHCl_3 = 25.5 g /119.5 g mol ^{-1} = 0.213 molTotal number of moles = 0.47 + 0.213 = 0.683 molMole fraction of component 2 = 0.47 mol / 0.683 mol = 0.688 Mole fraction of component 1 = 1.00 – 0.688 = 0.312 We know that: P_{T} = p_1^0+( p_2^0- p_1^0)x_2 = 200 + (415 – 200) \times 0.688 = 200 + 147.9 = 347.9 mm Hg

Q: A solution containing 0.5126 g naphthalene (mol. mass =128 ) in 50 g of carbon tetrachloride yield a boiling point elevation 0.402^0 C while a solution of 0.6216 g of an unknown solute in the same weight of the solvent gives a boiling elevation of 0.647^0C . Find the molecular mass of the unknown solute.

In this case, the first data is used to find the value of K_b .This value of K_b is used in the second data to find the molecular mass.Step I. To find K_b for CCl_4 from data on naphthalene solution w_2=0.5126\,g w_1=50.00\,g M_2=128\,g\,mol^{-1} \Delta T_b=0.402^0C=0.402\,K K_b=\dfrac{\Delta T_b.w_1M_2}{1000w_2}=\dfrac{0.402K\times 50g\times 128\,g\,mol^{-1}}{1000g\,kg^{-1}\times 0.5126g}=5.02\,K\,kg\,mol^{-1} Step II. To find the mol mass of unknown solute.The given data are: w_2=0.6216g,w_1=50.00g \Delta T_b=0.647^0C=0.647K K_b=5.02\,K\,kg\,mol^{-1} (calculate above) M_2=\dfrac{1000K_bw_2}{w_1\Delta T_b}=\dfrac{1000\,g\,kg^{-1}\times 5.02\,K\,kg\,mol^{-1}\times 0.6216g}{50g\times 0.647K}=96.46\,g\,mol^{-1}

Q: The Van't Hoff factor for 0.1 M La (NO_3)_3 solution is found to be 2.74 the percentage dissociation of the salt is :

Let x be the percentage dissociation. La(NO_3)_3 \rightarrow La^{3+} + 3 {NO}^{-}_{3} Initial moles 1 0 0 Moles after dissociation 1-x x 3xThe Van't Hoff factor is 2.74 i=\dfrac {Actual \ number \ of \ moles}{expected \ number \ of \ moles} 2.74 = \dfrac {1+3\alpha}{1} \alpha=0.58 \alpha \%=58\ \%

Q: There are KI and sucrose solutions with 0.1 M concentration each. If the osmotic pressures of KI and sucrose solutions are 0.465 bar and 0.245 bas respectively, find the degree of dissociation for KI.

For KI solution, \pi=i\,CRT \therefore 0.465=i\times 0.1\times R\times T For sucrose solution, \pi=CRT \therefore 0.245=0.1\times R\times T Dividing (i) by (ii), we get i=\dfrac{0.465}{0.245}=1.898 If \alpha is the degree of dissociation KI\,\,\rightleftharpoons\,\, K^++I^- Initial 1 moleAfter dissociation, 1-\alpha \alpha \alpha Total moles after dissociation =(1-\alpha)+\alpha+\alpha=1+\alpha \therefore i=1+\alpha or \alpha=i-1=1.898-1=0.898=89.8\%

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Problem solving tips

3 min read

- Tricks and Tips to solve difficult problem easily

Tip: Follow these simple tips to find the total vapour pressure over the solution phase from the given data:
Say you are given two liquids and both are volatile. When we mix the two liquids, both the liquids will eventually evaporate and exert a pressure over the solution phase.

To make things easier, we are already given the values of the vapour pressure of the pure components. Plus, we are given the amounts of the two components which have been mixed to form an ideal solution.

Step 1- First of all, we will find the number of moles of each component

For this, we will find the molar mass of each component, and then we can find the number of moles of each component by applying this expression

n = \frac{W}{M}

Here, W is the Weight of the component, and M is Molar mass of the component

Step 2- Once we get the values of number of moles of component 1 and component 2, we can find the mole fraction of component 1 and component 2

x_{A} = \frac{n _{A}}{n _{A} + n _{B}}

x_{B} = 1 - x_{A}

Step 3 -Now, go on and apply Raoult's law to find the partial vapour pressure of each component:

p_{A} = x_{A} p_{A}^{0}

(Equation 1)

p_{B} = x_{B} p_{B}^{0}

(Equation 2)

Step 4 -Finally, add equations 1 and 2 to find the total pressure:

p_{t o t a l} = p_{A} + p_{B} = x_{A} p_{A}^{0} + x_{B} p_{B}^{0}

Now, let's practice this question to become a pro!

Vapour pressure of chloroform

(C H C l_{3})

and dichloromethane

(C H_{2} C l_{2})

2 5^{o} C

are 200 mm Hg and 415 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of

C H C l_{3}

and 40 g of

C H_{2} C l_{2}

at the same temperature will be:

[Molecular mass of

C H C l_{3} = 119.5 g / m o l

and molecular mass of

C H_{2} C l_{2} = 85 g / m o l

]

615.0 mm Hg

347.9 mm Hg

285.5 mm Hg

173.9 mm Hg

View solution

Tip: I hope by now you have studied all the colligative properties of a solution. One of these properties include elevation in boiling point. Can you solve related questions easily? Yes?

Let's imagine two solutions with the same solvent but different solute particles.

To the first, we have added a given amount of a known solute - it's molecular mass is

M_{2}

. And the second contains a given amount of an unknown solute with molecular mass

M_{2}^{'}

. Additionally, we know the value of boiling point elevation in both the scenarios.

Now, can you find the value of molecular mass of the unknown solute?

In such questions you need to use the formula very smartly!

There are a few simple steps to follow-

Step 1- First of all, we will find the value of

K_{b}

using the values given for Solution 1. For this, we will use the expression-

K_{b} = \frac{Δ T _{b} . w _{1} M _{2}}{1 0 0 0 w _{2}}

Remember that the value of

K_{b}

is constant for a solvent.

Step 2- Now, we can place the value of

K_{b}

in the relevant expression to find the molecular mass of the unknown solute.

M_{2}^{'} = \frac{1 0 0 0 K _{b} w _{2}^{'}}{w _{1} Δ T _{b}}

Practice the question to get the better clarification.

A solution containing

0.5126 g

naphthalene (mol. mass

= 128

) in

50 g

of carbon tetrachloride yield a boiling point elevation

0.40 2^{0}

C while a solution of

0.6216 g

of an unknown solute in the same weight of the solvent gives a boiling elevation of

0.64 7^{0} C

. Find the molecular mass of the unknown solute.

View solution

Tip: Let's quickly revise the steps to find the degree of dissociation/association
Step 1 - Write the dissociation/association equation. Say, "a" mole of a solute is added to a solvent to make a solution. Let us assume

α

to be the degree of dissociation/association of the solute in the solvent.

	$A B$	$⇌$	$A^{+}$	$+$	$B^{-}$
Initially	a mole
After dissociation	$a - a α$		$a α$		$a α$

Step 2-Calculate van't hoff factor using the relation

i = \frac{Total no. of moles of particles after association or dissociation}{Number of moles of particles before association or dissociation}

Step3- Finally, calculate

α

by combining steps 1 and 2
Practice these questions by applying the above steps-

The Van't Hoff factor for 0.1 M La

(N O_{3})_{3}

solution is found to be 2.74 the percentage dissociation of the salt is :

85%

58%

65.8%

52%

View solution

There are KI and sucrose solutions with 0.1 M concentration each. If the osmotic pressures of KI and sucrose solutions are 0.465 bar and 0.245 bas respectively, find the degree of dissociation for KI.

0.987

0.623

0.747

0.897

View solution