Question #c1433

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Answer 1 · 2016-11-23T14:44:37

ATOMIC MASSES OF

$N->14" g/mol"$

$O->16" g/mol"$

Molar mass of $N_2O_3=2xx14+3xx16=76" g/mol"$

We know 1mol of any element or compound contains $6.023xx10^23$ molecules.

So $4.3xx10^21$ molecules of $N_2O_3$ will be

#(4.3xx10^21)/(6.023xx10^23)mole

$=4.3/6.023xx10^-2mol$

$=4.3/6.023xx10^-2molxx76g/"mol"$

$=0.5426g$