It is desired to prepare 500g of 10% mass by mass percentage of urea in water. How much urea should be dissolved in how much volume of water? Density of water is 1 g (mL) raised to the power -1.
It is desired to prepare 500g of 10% mass by mass percentage of urea in water. How much urea should be dissolved in how much volume of water? Density of water is 1 g (mL) raised to the power -1.
The mass by mass percentage of a solution is given by the following relation
mass by mass percentage of a solution = (mass of solute / mass of solution) X 100
where mass of the solution = mass of solute + mass of solvent
We have to prepare 500g of 10% mass by mass percent solution of urea in water. Let the mass of urea required be 'a' grams. Therefore
(a / 500) X 100 = 10
Solving for a, we get
a = 50g.
Thus, mass of urea is 50g. So, mass of water will be
= mass of solution - mass of urea
= (500 - 50)g = 450g
Since the density of water is given to be 1gml-1, therefore the volume of water required would be
= mass of water / density of water
= 450 / 1 = 450 ml.
Hence the mass of urea required would be 50g and the volume of water required would be 450 ml.
The mass by mass percentage of a solution is given by the following relation
mass by mass percentage of a solution = (mass of solute / mass of solution) X 100
where mass of the solution = mass of solute + mass of solvent
We have to prepare 500g of 10% mass by mass percent solution of urea in water. Let the mass of urea required be 'a' grams. Therefore
(a / 500) X 100 = 10
Solving for a, we get
a = 50g.
Thus, mass of urea is 50g. So, mass of water will be
= mass of solution - mass of urea
= (500 - 50)g = 450g
Since the density of water is given to be 1gml-1, therefore the volume of water required would be
= mass of water / density of water
= 450 / 1 = 450 ml.
Hence the mass of urea required would be 50g and the volume of water required would be 450 ml.
