Question

The number of molecules present in $2$ grams of helium at ${273}^{o}C$ and $1520$ $mmHg$ pressure are:

[$N_A=6.02\times {10}^{23}$]

• A. $3.01$ x ${10}^{23}$
• B. $6.02$ x ${10}^{23}$
• C. $3.01$ x ${10}^{22}$
• D. $6.02$ x ${10}^{22}$

Answer 1

The correct option is A $3.01$ x ${10}^{23}$

$4$ grams of $He$ (1 mole) contain $6.023\times {10}^{23}$ He molecules.

$2$ grams of $He$ contains $x$ molecules of $He$.

$x=\frac{6.023\times {10}^{23}}{2}=3.01\times {10}^{23}molecules$.

Hence, option A is correct.