Question

Calculate the mass of Urea (${ NH }_{ 2 }CO{ NH }_{ 2 }$) required in making 2.5 kg of 0.2 molal aqueous solutions.

Answer 1

Hint: We are given with total weight of the solution and molality of the solution. Let us assume some weight off for urea and calculate moles of urea we can get the weight of solvent from subtracting urea weight from the total weight. Equating calculated molality to given molality we get required weight.

Complete step by step solution:
We know that the molality of a substance is calculated as the number of moles of solute per kg of solvent. We first assume the weight of urea as ‘x’ grams and calculate the number of moles.
\[\]Number of moles of urea = $\dfrac { Weight }{ Molecular\quad weight\quad of\quad urea } $
\[\]The molecular weight of ${ NH }_{ 2 }CO{ NH }_{ 2 }\quad =\quad 2\quad \times \quad 14\quad +\quad 4\quad \times \quad 1\quad +\quad 12\quad +\quad 16\quad =\quad 28\quad +\quad 4\quad +\quad 28\quad =\quad 60$
\[\]Number of moles of urea = $\dfrac { x }{ 60 } $
\[\]The molality of urea = $\dfrac { Number\quad of\quad moles\quad of\quad urea }{ Weight\quad of\quad solvent\quad in\quad kg }$
\[\]Weight of solvent = $Weight\quad of\quad solvent\quad in\quad kg\quad =\quad 2.5\quad -\quad \dfrac { x }{ 1000 } \quad kg$ since x is in grams.
The molality of urea is given as 0.2.
\[0.2\quad =\quad \dfrac { x }{ 60 } \quad \times \quad \dfrac { 1 }{ 2.5\quad -\quad \dfrac { x }{ 1000 } }\]
\[0.2\quad \times \quad 60\quad \times \quad \left( 2.5\quad -\quad \dfrac { x }{ 1000 } \right) \quad =\quad x\]
\[2\quad \times \quad 6\quad \times \quad \left( 2.5\quad -\quad \dfrac { x }{ 1000 } \right) \quad =\quad x\]
\[12\quad \times \quad 2.5\quad -\quad \dfrac { 12x }{ 1000 } \quad =\quad x\]
\[30\quad =\quad \dfrac { 988x }{ 1000 }\]
\[x\quad =\quad \dfrac { 30000 }{ 988 } \quad =\quad 30.364\quad g\]

Therefore the weight of urea required is 30.364 grams.

Note: While calculating molality we need to be careful with the formula. We need to check whether we are taking the weight of the entire solution or only solvent. We need to be careful with the units in every step.