Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • When  of <img alt="\mathrm{ 0.5 \mathrm\; {M}}" src="/latex-image/?%5Cmathrm%7B%200.5%

When \mathrm{800 \mathrm{~mL}} of \mathrm{ 0.5 \mathrm\; {M}} nitric acid is heated in a beaker, its volume is reduced to half and \mathrm{ 11.5 \mathrm{~g}} of nitric acid is evaporated. The molarity of the remaining nitric acid solution is \mathrm{ x \times 10^{-2} \mathrm{M}. } (Nearest Integer)

\mathrm{ \text { (Molar mass of nitric acid is } 63 \mathrm{~g} \mathrm{~mol}^{-1} \text { ) }}

Option: 1

54


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Initially Solution is 0.5 m 800 ml \mathrm{HNO_{3}}

\mathrm{M\; HNO_{3}}=0.5\times 0.8\text{L = 0.4 mole}

\mathrm{W\; HNO_{3}}=0.4\times 63\text{ g/mole = 25.2 g }

On heating

Valume reduced to 400 ml and 11.5 g of \mathrm{HNO_{3}} is evaporated.

Wt.of \mathrm{HNO_{3}=25.2\; g-11.5\; g=13.7 \; g}

\mathrm{n\; HNO_{3}=\frac{13.7}{63\; g/mole}=0.21\; mole}

\mathrm{M\; HNO_{3}=\frac{0.217}{0.4\; L}=0.543 \; M}

                                        \mathrm{=54.3 \times 10^{-2}\; m}

Answer = 54

Posted by

vinayak

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