How many grammes of methyl acetylene (propyne) CH 3 -C=CH will completely discharge the colour...

Explanation

To answer this question, we need to understand the reaction between methyl acetylene (propyne) and bromine. Methyl acetylene, CH\(_3\)-C=CH, is an alkyne and reacts with bromine (Br\(_2\) in an addition reaction, where the triple bond in the alkyne breaks and two bromine atoms are added to the carbon atoms. The balanced equation for this reaction is:

\(CH_3CCH + Br_2 \rightarrow CH_3CHBrCHBr\)

Now, we need to calculate the number of moles of bromine present in 8 grams:

\(\frac{8 \text{ g}}{160 \text{ g mol}^{-1}} = 0.05 \text{ mol}\)

According to the balanced equation, one mole of methyl acetylene reacts with one mole of bromine. Therefore, we need 0.05 moles of methyl acetylene to completely discharge the colour of 8 grams of bromine.

To find the required grams of methyl acetylene, we can use its molar mass and the number of moles:

\(0.05 \text{ mol} \times (12 \times 3 + 1 \times 4) \text{ g mol}^{-1} = 0.05 \text{ mol} \times 40 \text{ g mol}^{-1} = 2 \text{ g}\)

So, 2 grams of methyl acetylene (propyne) are needed to completely discharge the colour of 8 grams of bromine, which corresponds toOption C: 2.0.