Question

How many grams of ch30h should be added to aqueous solution to prepare 150 ml solution of 2M ch30h

Answer 1

Let ‘x’ gram CH3OH is required,

We know that,

Molar mass of CH3OH = 32 grams

So number of moles  = Given weight / Molecular weight = x / 32.

Thus, molarity = Number of moles of solute / Volume of solution in Liters

Given molarity = 2

Therefore, 2 = $\frac{x / 32}{150}$  = 9.6g.

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