Question

The number of moles of $19.86$ g of $Pb(N{O}_3{)}_2$ is:

$[Pb=207,N=14,O=16]$.

• A. 0.6
• B. 0.06
• C. 6
• D. 60

Answer 1

The correct option is B 0.06

Let us consider the problem.

1 mol of $Pb{\left(N{O}_3\right)}_2=207+\left(14\times 2\right)+\left(16\times 6\right)=331gm$

In other word,

331 gram of $Pb{\left(N{O}_3\right)}_2$ is $=1mole$

Therefore, 19.86 gram is $1\times \frac{19}{331}=0.06moles$