Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

The higher oxide of an element (E) has the formula EO3. Its hydride contains 2.47% hydrogen.The element is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

S

b

Se

c

Si

d

Te

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

detailed_solution_thumbnail

Higher oxide of an element E is EO3. so oxidation state of E = +6. as the element has six electrons in its outer shell, element belongs to 16th group. The hydride of group 16 (S, Se, Te) have general formula H2E.

now 2.47% hydrogen contains in hydride of given element.

so, in 100g of hydride, contains 2.47 g hydrogen.

weight of hydrogen = 2.47

and then weight of element = 100 - 2.47 = 97.53g

now equivalents of element = equivalents of hydrogen

⇒97.53/(atomic weight/2) = 2.47/1

⇒97.53 × 2/2.47 = atomic weight of E

⇒atomic weight of E = 78.97 g/mol.

Therefore the element is Se (atomic weight of Se is 78.97 g/mol).

Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring