A commercially sold conc. HCl is 35% HCl by mass. If the density of this commercial acid is 1.46 g/mL, the molarity of this solution is

Option 1 - 10.2 M

Option 2 - 12.5 M

Option 3 - 14.0 M

Option 4 - 18.2 M

4 Views|Posted a year ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Correct Option - 3

Detailed Solution:

$M o l a l i t y = \frac{m o l e s o f s o l u t e * 1000}{w t o f s o l v e n t (g m)}$

$M o l a l i t y = \frac{35 * 1000 * 1.46}{36.5 * 100}$ = 14.0 M

Upvote

Similar Questions for you

CNG is an important transportation fuel. When 100g CNG is mixed with 208g oxygen in vechicles, it leads to the formation of CO₂ and H₂O and produces large quantity of heat during this combustion, then the amount of carbon dioxide produced in germs is ______.

(nearest integer)

[Assume CNG to be methane]

Vishal Baghel

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$

Mole $\frac{100}{16} \frac{208}{32}$

$\begin{array}{l} ↓ ↓ \\ 6.25 6.5 (H e r e O_{2} i s t h e l i m i t i n g r e a g e n t) \end{array}$

Hence mole of CO₂ formed = $\frac{6.5}{2}$

And wt of CO₂ in gm = $\frac{6.5}{2} \times 44 = 143 g$

The number of radial land angular nodes in 4d orbital are, respectively

Vishal Baghel

In 4d orbital, n = 4 and $l = 2$

Radial nodes = $n - l - 1$

Radial nodes = 4 – 2 – 1 = 1

And angular nodes, $l = 2$

Chlorophyll extracted form the crushed green leaves was dissolved in water to make 2 L solution of Mg of concentration 48 ppm. The number of atoms of Mg in this solution is x × 10²⁰ atoms. The value of x is _____________. (Nearest Integer)

(Given : Atomic mass of Mg is 24 g mol^-1; N_A = 6.02 × 10²³ mol^-1)

Vishal Baghel

Mass of solute = 48 * 2 * 10^-3 gram

Atoms of Mg = 24.08 * 10²⁰

A sample of 0.125g of an organic compound when analyzed by Duma’s method yields 22.78 mL of nitrogen gas collected over KOH solution at 280 K and 759 mm Hg. The percentage of nitrogen in the given organic compound is__________. (Nearest integer)

[Given: (a) The vapour pressure of water of 280K is 14.2mm Hg

(b) R = 0.082 L atm K^-1 mol^-1]

Vishal Baghel

$V_{N_{2}} = 22.78 m l T = 280 K$

PTotal = 759 mm of Hg

$P_{N_{2}} = 759 - 14.2 = 744.8 m m H g$

$n_{N_{2}} = \frac{744.8 \times (22.78 / 1000)}{(0.082 \times 760) \times 280}$

= 0.00097

$% N = \frac{0.02716}{0.125} \times 100 = 21.728$

$\approx 22 .$

2L of 0.2 M H₂SO₄ is reacted with 2L of 0.1 M NaOH solution, the molarity of the resulting product Na₂SO₄ in the solution is_________ millimolar (Nearest integer)

Vishal Baghel

H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O

Number of mole = 0.2 × 2 2 × 0.1 &nb

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

67K

Colleges

1.2K

Exams

7.2L

Reviews

1.9M

Answers

Learn more about...

Chemistry NCERT Exemplar Solutions Class 12th Chapter One 2025

View Exam Details

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

See what others like you are asking & answering