Question

100gm of an aqueous solution of sugar contains 40% sugar by mass. How much water should be evaporated to get 50% of sugar solution by mass?
(A) 10g
(B) 50g
(C) 0.0g
(D) 40g

Answer 1

Hint: The quantitative study of the reactants and products involved in a chemical reaction is known as chemical stoichiometry, which is generally a concern with numbers as the calculation of products and reactants. This concept in chemistry also helps to balance chemical equations.

Complete step by step solution:
According to the stoichiometry concept, to determine the amount of substance present or required for a chemical reaction can be measured are,
1. Reactants and products mass
2. Molecular weight
3. Chemical equations
4. Formulas
-Stoichiometric coefficient: The number of molecules involved or participates in the reaction is known as the stoichiometric number of the stoichiometric coefficient. In a balanced chemical reaction, there is an equal number of elements on both sides of the equation. This stoichiometric coefficient is a number represented in front of atoms, molecules, or ions.
-Given, 100gm of an aqueous solution of sugar contains 40% sugar by mass.
-100 gm aqueous solution with 40% sugar = 40 gm of sugar + 60 g water in solution.
-50 % sugar by mass added in 100gm of an aqueous solution presents the remaining 50g water.
-100 gm of aqueous solution with 50 % sugar = 50 g of sugar + 50 g water in solution.
-Amount of water evaporated to get 50% of sugar in solution = 60g – 50g =10g.
-Hence, 10g of water should be evaporated to get 50% of sugar solution by mass.

So, the correct answer is option A.

Note: The atoms and molecules are very small in size and their numbers in a very small amount of substance are large. Then the mole concept is introduced to represent atoms and molecules in bulk. One mole of a substance contains $6.023\times {{10}^{23}}$ a number of substances, which is known as Avogadro’s number. The atomic or formula mass is equal to the mass of one mole of a substance, which is known as molar mass.