QJEEM21Che6a9: JEE Main 2021 Chemistry - Class 11 JEEM21Che6a9 Solution: 4.5 g of compound A (MW = 90) was used to make 250 mL of its…

Given, weight of compound A = 4.5 g

Molecular weight of compound A = 90 g/mol

Volume of solution (in mL) = 250 mL

Now, molarity is defined as number of moles of solute or compound A divided by volume of solution (in L).

$M = {{Number\,of\,moles\,of\,solute\,(n)} \over {Volume\,of\,solution}}$

$= {{{{4.5} \over {90}}} \over {{{250} \over {1000}}}} = 0.2$ = 2 $\times$ 10 ^{$-$ 1} M

$\therefore$ $n = {{Weight\,of\,solute\,(compound\,A)} \over {Molecular\,weight\,of\,solute\,(compound\,A)}}$

Hence, x $\times$ 10 ^{$-$ 1} $\mu$

x = 2