CBSE Class 11-science Answered

Pls solve it.

Asked by pb_ckt | 01 May, 2019, 10:43: PM

Expert Answer

Option (b) is correct.

Given:

1 gm of hydrogen atoms

We know,

1 gm of hydrogen atoms = 1 mole

1 mole of hydrogen =6.022×10²³ atoms

1 atom of hydrogen conatains 1 electron

Therefore 1 gm hydrogen atoms contains 6.022×10²³ electrons.

Now, for 1 gm of oxygen atoms

No. of moles of oxygen

equals 1 over 16 equals space 0.0625 space mol

1 mol of oxygen atoms = 6.022×10²³ atoms

so 0.0625 mol f oxygen atoms =

fraction numerator 0.0625 cross times space 6.022 cross times 10 to the power of 23 space end exponent over denominator 1 end fraction equals space 3.76 space cross times 10 to the power of 22 space end exponent space atoms 1 space atom space of space oxygen space conatains space 8 space electrons space therefore space 3.76 space cross times 10 to the power of 22 space end exponent space atoms space will space contain space space fraction numerator 3.76 space cross times 10 to the power of 22 space end exponent space cross times 8 over denominator 1 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 3.11 cross times cross times 10 to the power of 23 space end exponent space electrons

For 2 gm oxygen atoms

No. of moles of oxygen

$equals 2 over 16 equals space 0.125 space mol$

1 mol of oxygen atoms = 6.022×10²³ atoms

so 0.125 mol f oxygen atoms =

$fraction numerator 0.125 cross times space 6.022 cross times 10 to the power of 23 space end exponent over denominator 1 end fraction equals 7.5 space cross times 10 to the power of 22 space end exponent space atoms 1 space atom space of space oxygen space conatains space 8 space electrons space therefore space 7.5 space cross times 10 to the power of 22 space end exponent space space atoms space will space contain space space fraction numerator 7.5 space cross times 10 to the power of 22 space end exponent space space cross times 8 over denominator 1 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 6.022 cross times 10 to the power of 23 space end exponent space electrons$