A) 8 g of \[\text{C}{{\text{H}}_{\text{4}}}\]
B) 4.4 g of \[\text{C}{{\text{O}}_{\text{2}}}\]
C) 34.2 g of\[{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\]
D) \[\text{2 g of }{{\text{H}}_{2}}\]
Correct Answer: D
Solution :
Number of particles in 8 g of \[\text{C}{{\text{H}}_{\text{4}}}\] \[\text{=}\frac{6.022\times {{10}^{23}}}{16}\times 8=3.0115\times {{10}^{23}}\] No. of particles in 4.4 g of \[\text{C}{{\text{O}}_{\text{2}}}\] \[\text{=}\frac{6.022\times {{10}^{23}}}{44}\times 4.4=6.22\times {{10}^{22}}\] No. of particles in 34.2 g of \[{{\text{C}}_{\text{12}}}{{\text{H}}_{\text{22}}}{{\text{O}}_{\text{11}}}\] \[\text{=}\frac{6.022\times {{10}^{23}}}{342}\times 34.2=6.022\times {{10}^{22}}\] No. of particles in 2 g of \[{{\text{H}}_{\text{2}}}\] \[\text{=}\frac{6.022\times {{10}^{23}}}{2}\times 2=6.022\times {{10}^{23}}\]You need to login to perform this action.
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