Question
Question: Calculate the amount of benzoic acid \( \left( {{C}_{6}}{{H}_{6}}COOH \right) \) required for prepar...
Calculate the amount of benzoic acid (C6H6COOH) required for preparing 250 mL of 0.15M solution in methanol.
(A) 0.4575g
(B) 4.575g
(C) 45.75g
(D) 0.04575g
Solution
Hint : We know that to solve this question, we must first understand the whole conversion of Ethanol into Methanol. Then we need to observe the reagents used in the reaction and then only we can have our correct answer.
Complete Step By Step Answer:
Before we move forward with the solution of this given question, let us first understand some basic concepts about ethanol: Ethanol, also called alcohol, ethyl alcohol and grain alcohol, is a clear, colorless liquid and the principal ingredient in alcoholic beverages like beer, wine or brandy. Because it can readily dissolve in water and other organic compounds, ethanol also is an ingredient in a range of products, from personal care and beauty products to paints and varnishes to fuel.
Number of moles of benzoic acid required =0.15 ×1000250=0.0375 moles
Molar mass of benzoic acid =7(12)+6(1)+2(16)=84+6+32=122 g/mol
Mass of benzoic acid required =122×0.0375=4.575 g
In such types of problems, we consider that all of the solute dissolves completely in the solution but in reality the situation may be different due to various factors like temperature, saturation, pressure, etc. Benzoic acid dissolves readily in methanol (solubility of benzoic acid at −18∘C in methanol is 30g/100g ).
Therefore, the correct answer is option B.
Note :
Remember that Methanol is a non-drinking type of alcohol (also known as wood alcohol and methyl alcohol) which is mostly used to create fuel, solvents and antifreeze. A colorless liquid, it is volatile, flammable and unlike ethanol, poisonous for human consumption. Methanol is also used to produce a variety of other chemicals, including acetic acid.