Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A commercially sold conc. \mathrm{HCl}$ is $35 \% \mathrm{HCl}  by mass. If the density of this commercial acid is \mathrm{1.46\: g/mL,}  the molarity of this solution is :
(Atomic mass : \mathrm{Cl}=35.5\, \mathrm{amu}, \mathrm{H}=1 \mathrm{amu} )

Option: 1

10.2 \mathrm{~M}


Option: 2

12.5 \mathrm{~M}


Option: 3

14.0 \mathrm{~M}


Option: 4

18.2 \mathrm{~M}


Answers (1)

best_answer

Given,

Concentration of \mathrm{HCl}=35 \% \mathrm{HCl} by mass.
\mathrm{density =1.46 \mathrm{~g} / \mathrm{ml}}

Let us assume the total volume of Solution \mathrm{\mathrm{=1000\, \mathrm{ml}}}

From density, total mass of Solution \mathrm{=1.46 \times 1000=1460 ~g}

\mathrm{\therefore\: mass \: of\: \mathrm{HCl}=\frac{35}{100} \times 1460=511~g}

\mathrm{\text{numeber of moles of }\: \mathrm{HCl}=\frac{\text { mass of } \mathrm{HCl} \text { in Solution }}{\text { molar mass of HCl }}}
                                                         \mathrm{=\frac{511}{36.5}=14}

\mathrm{Molarity = \text{no. of moles of}\: \mathrm{HCl} \: in \: 1000 \mathrm{ml}\: \: of \: solution}
                    \mathrm{=14 {~M}}

Hence, option (C) is correct.

Posted by

manish

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: 11 from Kota coaching centres among 24 students with 100 percentile
anam-khan

Low JEE Main rank? Explore alternative paths for engineering admission
anam-khan

Gender-gap in JEE Main 2025: Female participation remains around 31%, overall pool drops by 7.09%

Latest Question