Question

|⭕+QUESTION+⭕|

2ᵍ ᵒᶠ ᵇᵉⁿᶻᵒⁱᶜ ᵃᶜⁱᵈ (ᶜ6ʰ5ᶜᵒᵒʰ) ᵈⁱˢˢᵒˡᵛᵉᵈ ⁱⁿ 25ᵍ ᵒᶠ ᵇᵉⁿᶻᵉⁿᵉ ˢʰᵒʷˢ ᵃ ᵈᵉᵖʳᵉˢˢⁱᵒⁿ ⁱⁿ ᶠʳᵉᵉᶻⁱⁿᵍ ᵖᵒⁱⁿᵗ ᵉQ꙰ᵘᵃˡ ᵗᵒ 1.62 ᵏ. ᵐᵒˡᵃʳ ᵈᵉᵖʳᵉˢˢⁱᵒⁿ ᶜᵒⁿˢᵗᵃⁿᵗ ᶠᵒʳ ᵇᵉⁿᶻᵉⁿᵉ ⁱˢ 4.9 ᵏ ᵏᵍ/ᵐᵒˡ. ʷʰᵃᵗ ⁱˢ ᵗʰᵉ ᵈᵉᵖʳᵉˢˢⁱᵒⁿ ᵃˢˢᵒᶜⁱᵃᵗⁱᵒⁿ ᵒᶠ ᵃᶜⁱᵈ ⁱᵗ ⁱᶠ ᶠᵒʳᵐˢ ᵈⁱᵐᵉʳ ⁱⁿ ˢᵒˡᵘᵗⁱᵒⁿ?​

Answer 1

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\bold{QUESTION:}$

2ᵍ ᵒᶠ ᵇᵉⁿᶻᵒⁱᶜ ᵃᶜⁱᵈ (ᶜ6ʰ5ᶜᵒᵒʰ) ᵈⁱˢˢᵒˡᵛᵉᵈ ⁱⁿ 25ᵍ ᵒᶠ ᵇᵉⁿᶻᵉⁿᵉ ˢʰᵒʷˢ ᵃ ᵈᵉᵖʳᵉˢˢⁱᵒⁿ ⁱⁿ ᶠʳᵉᵉᶻⁱⁿᵍ ᵖᵒⁱⁿᵗ ᵉQ꙰ᵘᵃˡ ᵗᵒ 1.62 ᵏ. ᵐᵒˡᵃʳ ᵈᵉᵖʳᵉˢˢⁱᵒⁿ ᶜᵒⁿˢᵗᵃⁿᵗ ᶠᵒʳ ᵇᵉⁿᶻᵉⁿᵉ ⁱˢ 4.9 ᵏ ᵏᵍ/ᵐᵒˡ. ʷʰᵃᵗ ⁱˢ ᵗʰᵉ ᵈᵉᵖʳᵉˢˢⁱᵒⁿ ᵃˢˢᵒᶜⁱᵃᵗⁱᵒⁿ ᵒᶠ ᵃᶜⁱᵈ ⁱᵗ ⁱᶠ ᶠᵒʳᵐˢ ᵈⁱᵐᵉʳ ⁱⁿ ˢᵒˡᵘᵗⁱᵒⁿ?

$\rule{300}{1.5}$

$\huge{\pink {\sf\underline{Answer}}}$

refer to the attachment!!

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$<font color =pink ><marquee behavior=alternate>Thanks </marquee></font>$

Answer 2

$\huge{\boxed{\bold\orange{\fcolorbox{red}{black}{Your\:Answer:}}}}$

▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪

▶given◀

W2 = 2 g

Kf = 4.9 K kg/mol

W1 = 25g

∆Tf = 1.62 K

________________________________

▶Solution◀

$m_{2} = \frac{4.9k \: kg \: {mol}^{ - 1} \times 2g \times 1000g \: {kg}^{ - 1} }{25g \times 1.62k} \\ = 241.98g \: {mol}^{ - 1}$

Thus, experimental molar mass of benzoic acid in benzene is

= 2.41.98g/mol

Now consider the following equilibrium for the acid:

2C6H6COOH -------> ( C6H5COOH)2

Thus, total number of molar of particles at equilibrium equals van't Hoff factor i.

$i = \frac{normal \: molar \: mass}{abnormal \: molar \: mass} \\ \\ = \frac{122g \: {mol}^{ - 1} }{241.98g \: {mol}^{ - 1} } \\ \\ \frac{x}{2} = 1 - \frac{122}{241.98} \\ \\ = 1 - 0.504 = 0.496 \\ \\ x = 2 \times 0.496 = 0.992$

Therefore, degree of association of benzonic acid in benzene is 99.2℅.

____________________________________

$<marquee></strong><strong>▶</strong><strong>hope</strong><strong> </strong><strong>it</strong><strong> </strong><strong>helps</strong><strong> </strong><strong>you</strong><strong>◀</strong><strong></</strong><strong>marquee</strong><strong>>$