Question

Calculate the mass of a non volatile solute which should be dissolved in 114g of octane to reduce its vp to 80%

Answer 1

☆Hey friend!!!! ☆

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here is your answer ☞
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Let the vapour pressureof pure octane be
$p _{1} ^{0}$


Then, the vapourpressure of the octane after dissolving the non-volatile solute is 
$\frac{80}{100} \: p {}^{0} _{1} = 0.8p _{1}{}^{0}$


Molar mass of solute,M2 = 40 g mol−1

Mass of octane, w1= 114 g

Molar mass of octane,(C8H18), M1 = 8 × 12 +18 × 1

= 114 g mol−1

Applying the relation,

$= > \: \frac{p _{1} ^{0} - p _{1} }{p _{1} ^{0} } = \frac{w _{2} \times m _{1} }{m _{2} \times w _{1} } \\ \\ = > \: \frac{ p _{1} ^{0} - 0.8p _{1} ^{0} }{p _{1} ^{0} } = \frac{w _{2} \times 114}{40 \times 114} \\ \\ = > \: w _{2} = 8 \: gm \: ...............answer$


Hence, the requiredmass of the solute is 8 g.


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hope it will help you ☺☺☺
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Devil_king ▄︻̷̿┻̿═━一

Answer 2

Answer:

the mass of a non-volatile solute that should be dissolved in 114g of octane to lower its vapour pressure by 20% is 10g.

Explanation:

According to Raoult's regulation, relative reducing of vapour stress,

p∘A−psp∘A=xB……

(i)xB=nBnB+nA=WBMBWBMB+WAMA……

(ii)

Given Vapour stress is reduced to eighty% whilst non-unstable solute is dissolved in octane. It approach

If p∘A=1 atm then ps=0.8 atm; p∘A−ps=zero.

2 atm; MA(C8H18)=114 g mol−

1;WA=114 g;MB=forty g mol−1;WB=?

making use of equation

(ii)

zero.

2=WB40WB40+114114=WB40WB40+10.2=WBWB+400.2WB+eight=WB

WB−0.2WB=8

0.eight WB=8

WB=eighty.eight

WB=10 grams

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