Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A \mathrm{42.12 \%(\mathrm{w} / \mathrm{v})} solution of \mathrm{NaCl} causes precipitation of a certain sol in 10 hours. The coagulating value of \mathrm{NaCl} for the sol is

\text { [Given: Molar mass : } \left.\mathrm{Na}=23.0 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{Cl}=35.5 \mathrm{~g} \mathrm{~mol}^{-1}\right]

Option: 1

\mathrm{36\; \mathrm{mmol}\; \mathrm{L}^{-1}}


Option: 2

\mathrm{36 \mathrm{~mol} \mathrm{~L}^{-1}}


Option: 3

1440 \mathrm{~mol} \mathrm{~L}^{-1}


Option: 4

1440 \, \mathrm{mmol} \mathrm{L}^{-1}


Answers (1)

best_answer

\mathrm{42.12 \%(w / \mathrm{v}) \mathrm{NaCl}}  solution means 

\mathrm{ 42.12 \mathrm{~g}\: \mathrm{NaCl} \text { is present in } 100 \mathrm{~mL} \text { solution }}

\therefore\mathrm{ \text { Molarity of NaCl solution } =\frac{42.12}{58.5 \times 100} \times 1000 \\ }\\

                                                                 \mathrm{=7.2 \mathrm{M}}

\mathrm{7.2 \: mol\: NaCl\: present\: in\: 1 litre\: solution} \\

\mathrm{7200 \: \mathrm{mmol} \: \mathrm{NaCl}\: present\: in \: 1\: litre\: solution}

Now, \mathrm{ 7200 \: \mathrm{mmol} /L\: \mathrm{NaCl}} causes precipitation of a certain solution in \mathrm{10\: hr}

\mathrm{NaCl} electrolyte in milli moles per litre required to cause precipitation of a sol in two hours  \mathrm{=\frac{7200}{10} \times 2} ~\mathrm{=1440\: \mathrm{mmol}\: \mathrm{L}^{-1}}

Hence correct answer is option 4

Posted by

admin

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 ECE cut-offs for female candidates go up at top NITs
anam-khan

JEE Main 2025 Session 2 City Slip (OUT) Live: Admit card likely on March 29 at jeemain.nta.nic.in; cut-offs
anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads

Latest Question