$(a)\;8.5g\qquad(b)\;0.85g\qquad(c)\;0.92g\qquad(d)\;0.085g$

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Gram-molecular mass of ammonia=17g

Number of molecules in 17g(one mole) of $NH_3$

$\Rightarrow 6.02\times 10^{23}$

Let the mass of 3.01$\times 10^{22}$ molecules of $NH_3$ be =xg

So,

$\large\frac{3.01\times 10^{22}}{6.03\times 10^{23}}=\frac{x}{17}$

$x=\large\frac{17\times 3.01\times 10^{22}}{6.02\times 10^{23}}$

$\Rightarrow 0.85g$

Hence (b) is the correct answer.

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