Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The vapour pressure of acetone at $25^{\circ}$ is 192 torr. When $1.5 \mathrm{~g}$ of non-volatile substance was dissolved in $100 \mathrm{~g}$ of acetone at $25^{\circ} \mathrm{C}$ its vapour pressure was 189 torr. The molar mass $\left(\mathrm{gmol}^{-1}\right)$ of substance is
Option: 1
$\mathrm{ 42=m }$

Option: 2
$\mathrm{ 92=m }$

Option: 3
$\mathrm{ 55.6=m }$

Option: 4
$\mathrm{102=m }$

Answers (1)

best_answer

$\mathrm{\frac{192-189}{192}=\frac{1.5 / m}{100 / 58} }$
$\mathrm{m=55.6 }$ .

Posted by

Kuldeep Maurya

View full answer

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NEET UG: Delhi HC seeks NTA's response on plea alleging defective biometric system at centre

anam-khan

‘NTA is national corruption agency; conduct high-level probe or dissolve': NSUI after CBI FIR, arrests in NEET

anam-khan

'Munna Bhai' again: CBI books two medical students for fraud in NEET-UG 2023

Latest Question