The Novolac polymer has mass of 963 g. The number of monomer units present in it are

Question

The Novolac polymer has mass of 963 g. The number of monomer units present in it are (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Monomer unit of Novolac is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/edf7d7701756d1111270478acb817c06-.png" /> its molecular mass is 124 amu. Upon considering molecular weight of polymer as 963 amu (In question its given as 963 gram) Now if during formation of Novolac, (n-1) unit of water are removed then n × 124=963+[18 ×( n -1)] n =9

Q. The Novolac polymer has mass of $963 g$ . The number of monomer units present in it are

Monomer unit of Novolac is

its
molecular mass is $124$ amu. Upon considering molecular weight of polymer as $963$ amu (In question its given as $963$ gram) Now if during formation of Novolac, $(n - 1)$ unit of water are removed then
$n \times 124 = 963 + [18 \times (n - 1)]$
$n = 9$

Q. The Novolac polymer has mass of 963g. The number of monomer units present in it are

Monomer unit of Novolac is its molecular mass is 124 amu. Upon considering molecular weight of polymer as 963 amu (In question its given as 963 gram) Now if during formation of Novolac, (n−1) unit of water are removed then n×124=963+[18×(n−1)] n=9

Q. The Novolac polymer has mass of $963 g$ . The number of monomer units present in it are

Monomer unit of Novolac is

its
molecular mass is $124$ amu. Upon considering molecular weight of polymer as $963$ amu (In question its given as $963$ gram) Now if during formation of Novolac, $(n - 1)$ unit of water are removed then
$n \times 124 = 963 + [18 \times (n - 1)]$
$n = 9$