Question

(111) Calculate
(a) number of moles in 12.044x 1023 number of CO2 molecules.
(Atomic mass: C=12u, O=16u).
(b) Mass of 0.5 mole of NH3 ( atomic mass: N=14u, H=1u)
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Answer 1

here is the answer...hope it helps

Answer 2

   

$(a)\ \ No.\ of\ moles\ = \frac{No.\ of\ particles}{N_A} \\ \\ \Rightarrow\ \frac{12.044 \times 10^{23}}{6.022 \times 10^{23}} \\ \\ \Rightarrow\ \bold{2\ mol} \\ \\ [ Atomic\ masses\ are\ not\ actually\ needed\ here.] \\ \\ \\ (b)\ \ GMM\ of\ NH_3 \\ \\ \Rightarrow\ 14+3 \times 1 \\ \\ \Rightarrow\ 14+3 \\ \\ \Rightarrow\ 17g \\ \\ \\ \ Mass\ in\ grams\ = No.\ of\ moles\ \times GMM \\ \\ \Rightarrow\ 0.5 \times 17 \\ \\ \Rightarrow\ \bold{8.5g}$

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