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Question

Chemistry Question on Some basic concepts of chemistry

4⋅484\cdot48 litre of methane at S.T.P.S.T.P. corresponds to

A

1⋅2×10221\cdot2 \times 10^{22} molecules of methane

B

0⋅50\cdot5 mole of methane

C

3⋅2 g3\cdot2\, g of methane

D

0⋅10\cdot1 mole of methane

Answer

1⋅2×10221\cdot2 \times 10^{22} molecules of methane

Explanation

Solution

4⋅484\cdot48 litre CH4=4⋅4822⋅4=15=0⋅2 molCH_{4}=\frac{4\cdot48}{22\cdot4}=\frac{1}{5}=0\cdot2\,mol =16×0⋅2=3⋅2 g CH4=16 \times 0\cdot2=3\cdot2\,g\,CH_{4} =0⋅2×6⋅02×1023=0\cdot2 \times 6\cdot02\times10^{23} =1⋅2×1022=1\cdot2 \times10^{22} molecules