Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is <img alt="0.0198^{circ} mathrm{C}" src="https://en

1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is 0.0198^{circ} mathrm{C}. The percentage of dissociation of the acid is __________.(Nearest integer)

[ Given :   Density of acetic acid is 1.02 mathrm{~g} mathrm{~mL}^{-1}

                 Molar mass of acetic acid is 60 mathrm{~g} mathrm{~mol}^{-1}

                 left.mathrm{K}_{f}left(mathrm{H}_{2} mathrm{O} ight)=1.85 mathrm{~K} mathrm{~kg} mathrm{~mol}^{-1} ight]

Option: 1

5


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{\text{mass of acetic acid} =1.02 \times 1.2 g} \mathrm{=1.224 \mathrm{~g}}

\therefore\text{moles of acetic acid} =0.0204

\therefore \text { molality }=\frac{0.0204}{2}=0.0102

Now,

\mathrm{\Delta T_{f}=i K_{f} m }\\

\mathrm{\Rightarrow 0.0198=i(1.85)(0.0102)}

\mathrm{\Rightarrow i=1.05} \\

\mathrm{\Rightarrow 1+\alpha=1.05} \\

\mathrm{\Rightarrow \alpha=0.05} \\

\mathrm{\Rightarrow \% ~\alpha=5}

Hence answer is 5

Posted by

HARSH KANKARIA

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: BTech CSE cut-offs for PwD candidates at NITs make entry easy; top institutes still selective
anam-khan

73 or 99 percentile? JEE Main result questioned, J-K candidate says, ‘Don’t create hype’
anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile

Latest Question