Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • When a certain amount of solid is dissolved in of water at

When a certain amount of solid \mathrm{A} is dissolved in 100 \mathrm{~g} of water at 25^{\circ} \mathrm{C} to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is 23.76 \mathrm{mm\: Hg}. The number of moles of solute \mathrm{A} added is__________. (Nearest Integer)

Option: 1

3


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

We know the formulas

\mathrm{\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{n_{2}}{n_{1}+n_{2}} \quad\left\{\begin{array}{l} n_{2}=\text { mole of solute } \\ n_{1}=\text { mole of solvent } \end{array}\right\}}

 due to dilute solution  \mathrm{n_{1} \gg n_{2}\; \; so, n_{1}+n_{2}=n_{1}}

then  \mathrm{\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{n_{2}}{n_{1}}, }

\mathrm{given \: p_{1}^{0}=23.76\: \mathrm{mmHg} }
\mathrm{and \: p_{1}=\frac{p_{1}^{0}}{2}}
So,
\mathrm{\frac{p_{1}^{0}-\frac{p_{1}^{0}}{2}}{p_{1}^{0}} =\frac{n_{2}}{n_{1}}}
\mathrm{\frac{1}{2} =\frac{n_{2}}{w_{0} / M_{1}}}
\mathrm{n_{2} =\frac{w_{1}}{M_{1}} \times \frac{1}{2}=\frac{100}{18 \times 2}}
\mathrm{n_{2} =2.77 \approx 3}

Ans = 3
 

Posted by

Gautam harsolia

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

Calculate JEE Main 2025 rank using predictor tool for admission to top IITs, NITs, IIITs
anam-khan

NTA drops 1 out of 9 flagged errors from JEE Mains final answer key 2025
anam-khan

Consumer protection watchdog warns coaching centres against misleading ads ahead of IIT-JEE Main results

Latest Question