Question

How many grams of methanol ${CH}_{3}OH$ must be added to $2$ moles of $H_{2}O$ to make a solution containing equal numbers of molcules of $H_{2}O$ and ${CH}_{3}OH$?

• A. $2$ grams
• B. $16$ grams
• C. $32$ grams
• D. $64$ grams

Answer 1

The correct option is D $64$ grams

1 mole of $C{H}_{3}OH$ is equal to 32 g.

So, 2 moles of $C{H}_{3}OH$ is equal to 32$\times$2=64 g.