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Questions  >>  SAT  >>  Chemistry  >>  Atomic Structure
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Q)

Find the total mass of protons in 34 mg of $NH_3$ at STP.

$\begin{array}{1 1}(a)\;2.0145 \times 10^{5} \; kg \\ (b) \; 1.0145 \times 10^{-5} \; kg \\ (c) \; 2.0145 \times 10^{-7} \; kg \\ (d) \; 2.0145 \times 10^{-5} \; kg\end{array}$

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A)
1 mol of NH3=17gNH3" role="presentation" style="position: relative;">NH3=17gNH3NH3=17gNH3NH_{3} = 17g \; NH_{3} = 6.022×1023" role="presentation" style="position: relative;">6.022×10236.022×10236.022 \times 10^{23} molecules of NH3" role="presentation" style="position: relative;">NH3NH3NH_{3}
= (6.022×1023)(7+3)" role="presentation" style="position: relative;">(6.022×1023)(7+3)(6.022×1023)(7+3)(6.022 \times 10^{23})(7 + 3) proton = 6.022×1024" role="presentation" style="position: relative;">6.022×10246.022×10246.022 \times 10^{24} protons
Therefore 34 mg i.e 0.034gNH3=6.022×1024×0.034=1.2044×1022" role="presentation" style="position: relative;">0.034gNH3=6.022×1024×0.034=1.2044×10220.034gNH3=6.022×1024×0.034=1.2044×10220.034g \; NH_{3} = \; 6.022\times 10^{24} \times 0.034 = 1.2044 \times 10^{22} protons
mass of 1 proton =1.6726×10−27" role="presentation" style="position: relative;">=1.6726×1027=1.6726×10−27= 1.6726 \times 10^{-27} kg
Therefore mass of 1.2044×1022" role="presentation" style="position: relative;">1.2044×10221.2044×10221.2044 \times 10^{22} protons = (1.6726×10−27)(1.2044×1022)kg=2.0145×10−5" role="presentation" style="position: relative;">(1.6726×1027)(1.2044×1022)kg=2.0145×105(1.6726×10−27)(1.2044×1022)kg=2.0145×10−5(1.6726\times 10^{-27})(1.2044\times 10^{22}) kg = 2.0145\times 10^{-5} kg
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