Question

Question is in attachment​

Answer 1

Question:

Density of 2M $CH_3COOH$ is 1.2 gm / mL.

Find

1. %w/w
2. m
3. xSolute

Solution:

Density= 1.2 gm / mL

Molarity = 2M

%w/v = 12 %

1)

$\sf Density = \frac{\%w/v}{\%w/w} \\\\\sf 1.2 = \frac{12}{\%w/w} \\\\\sf \%w/w = \frac{12}{1.2} \\\\\sf \%w/w = 10 \%$

2)

Since the density is 1.2 gm/ mL, 1L will contain 1200 gm of solute.

Number of moles of solute in 1 litre given is 2 moles [ Molarity is 2]

Weight of solute = Moles x molar mass = 60 x 2 = 120 gm

Weight of solvent = 1200 - 120 = 1080 gm

Moles of solvent: 18g water has 1 mole, 1080 gm has 60 moles

Total moles in solution is 60 + 2 = 62 moles

Calculate molality

$\sf molality = \frac{n Solute}{wSolvent} \\\\\sf molality = \frac{2 \times 1000}{1080} \: => \frac{50}{27} => 1.85 moles$

3)

Calculate xSolute i.e mole fraction of solute.

$\sf xSolute= \frac{nSolute}{nSolvent + nSolution} \\\\\sf xSolute = \frac{2}{62} = 1/31$

Answer 2

$\maltese \: \large\underbrace{ \pmb{ \rm{Solution \: 1 :- }}}$

• Density = 1.2 gm/mL

• Molarity = 2M

• %W/W = ?

Formula to be applied :-

$\\ \leadsto \underline{ \boxed{ \sf M = \dfrac{ \%\dfrac{w}{w} \times density _{(solution)} \times 10 }{ molecular \: wt_{(solute)}} }} \bigstar$

Where, M denotes Molarity.

$\dag \: \underline{ \sf substitute \: the \: values \: to \: get \: the \: required \: answer}$

$\\ \quad: \implies \sf 2 = \dfrac{\% \dfrac{w}{w} \times 1.2 \times 10 }{60}$

$\\ \quad: \implies \sf \% \frac{w}{w} = \frac{2 \times 60}{1.2 \times 10}$

$\\ \quad: \implies \boxed{\boxed{\sf \orange{\%\frac{w}{w} = 10\% }}} \bigstar$

___________________________

$\maltese \: \large\underbrace{ \pmb{ \rm{Solution \: 2 :- }}}$

In the previous question, we calculated the %W/W as 10% This means that, the total solute present in the solution is 10g and the total solvent present in it is 90g.

Hence, Molality will be calculated by the following formula:-

$\\ \leadsto \underline{ \boxed{ \sf m = \frac{ moles_{(solute)} }{wt \: of \: solvent} \times 1000}} \bigstar$

Where, m denotes Molality.

First, lets find out the moles of solute ...

$\\ \rightarrow \sf moles = \dfrac{given \: mass}{molar \: mass}$

$\\ \rightarrow \sf moles = \dfrac{10}{60}$

$\\ \therefore \sf moles = \frac{1}{6}$

$\dag \: \underline{ \sf substitute \: the \: values \: to \: get \: the \: required \: answer}$

$\\ \quad : \implies \sf m = \dfrac{ \dfrac{1}{6} }{90} \times 1000$

$\\ \quad : \implies \sf m = \frac{1}{6 \times 90} \times 100$

$\\ \quad : \implies \boxed{ \boxed{ \orange{\sf m = 1.85m}}} \bigstar$

____________________________

$\maltese \: \large\underbrace{ \pmb{ \rm{Solution \: 3 :- }}}$

It is provided that, Molarity of $\sf CH_{3}COOH$ is 2, this means that, 2 moles of solute is present in 1 L of solution. Now, Weight of Solute = 2 × 60 = 120 g

Also, density of the solution is 1.2 gm/mL . So, 1 L will contain 1200 g of solution.

Hence, weight of Solvent can be calculated as follows:-

• Wt of Solute + Wt of Solvent = Wt of Solution

• 120 + Wt of Solvent = 1200

• Wt of Solvent = 1080 g

Now, Mole fraction of $\sf CH_{3}COOH$ can be calculated by using the formula outlined as under :-

$\\ \leadsto \underline{ \boxed{ \sf X _{(solute)} = \frac{ moles_{(solute)} }{ mole_{(solute)} + moles_{(solvent)} } }} \bigstar$

Moles of Solvent = 1080/18 = 60

$\\ \quad : \implies \sf X _{(solute)} = \frac{2}{60 + 2}$

$\\ \quad : \implies \boxed{ \boxed{ \orange{\sf X _{(solute)} = \dfrac{1}{31} }}} \bigstar$

____________________________

hope its helpful :"D