Question

The density of pure water is:
(A) $ 0.1g/c{{m}^{3}}. $
(B) $ 1g/c{{m}^{3}}. $
(C) $ 100g/c{{m}^{3}}. $
(D) $ 1kg/c{{m}^{3}}. $

Answer 1

Hint: We know that the molality of the solution is one of the major factors in calculating the concentration of the solution. The factors which can be used to calculate the molality of the solution are the number of moles of the solute and the mass of the solvent in kg. With the help of the density of pure water, the number of moles of the water can be calculated.

Complete answer:
The concentration of the solution tells the strength of the solution. There are many factors that can be used for calculating the strength of the solution like molarity, molality, normality, mole fraction, ppm, etc. So, the molality of the solution is one of the major factors in calculating the concentration of the solution. There are two factors that are used for the calculation of the molality of the solution, i.e., the number of moles of the solute and the mass of the solvent in kg. When we divide the number of moles of the solute by the mass of the solvent in kg, we get the molality of the solution.
The density of pure water at room temperature i.e., is $ 0.9970749\text{ }g/mol $ , therefore the mass will be $ 0.9970749\text{ }kg, $ and the molecular mass is $ 18.0148\text{ }g/mol. $
Water never has an absolute density because its density varies with temperature. Water has its maximum density of $ 1g/c{{m}^{3}}~ $ or $ 1000\text{ }kg/{{m}^{3}}~ $ at $ 4 $ degrees Celsius.
Therefore the correct answer is option B.

Note:
Remember that since there is no volume in the calculation of the molality of the solution, so the molality is temperature independent and does not change even on changing the temperature of the solution.