Question Number 188775 by mustafazaheen last updated on 07/Mar/23
![how is solution 72.5gr of the [C_3 H_6 O] how many Molecule of [H] exist?](https://www.tinkutara.com/question/Q188775.png)
$${how}\:{is}\:{solution} \\ $$$$\mathrm{72}.\mathrm{5}{gr}\:\:\:{of}\:{the}\:\left[{C}_{\mathrm{3}} {H}_{\mathrm{6}} {O}\right]\:{how}\:{many}\:\mathrm{Molecule}\:{of}\:\left[{H}\right]\:{exist}? \\ $$
Answered by witcher3 last updated on 06/Mar/23
![M(C_3 H_6 O)=3.12+6.1+16=58g/Mol n=(m/M)=((72.5)/(58)) N=n.N_A =((72.5)/(58)).6,02.10^(23) ..Molecule of [C_3 H_6 O] atom of H=6.N](https://www.tinkutara.com/question/Q188778.png)
$$\mathrm{M}\left(\mathrm{C}_{\mathrm{3}} \mathrm{H}_{\mathrm{6}} \mathrm{O}\right)=\mathrm{3}.\mathrm{12}+\mathrm{6}.\mathrm{1}+\mathrm{16}=\mathrm{58g}/\mathrm{Mol} \\ $$$$\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}=\frac{\mathrm{72}.\mathrm{5}}{\mathrm{58}} \\ $$$$\mathrm{N}=\mathrm{n}.\mathrm{N}_{\mathrm{A}} =\frac{\mathrm{72}.\mathrm{5}}{\mathrm{58}}.\mathrm{6},\mathrm{02}.\mathrm{10}^{\mathrm{23}} ..\mathrm{Molecule}\:\mathrm{of}\:\left[\mathrm{C}_{\mathrm{3}} \mathrm{H}_{\mathrm{6}} \mathrm{O}\right] \\ $$$$\mathrm{atom}\:\mathrm{of}\:\mathrm{H}=\mathrm{6}.\mathrm{N} \\ $$