Question

It is desired to obtain the highest possible volume of 1.25 M urea, [Urea (aq)], from the following sources: 345 mL of 1.29 M urea solution, 485 mL of 0.653 M urea solution and 835 mL of 0.775 M urea solution. How can it be done? What is the maximum volume of solution that can be obtained?

Answer 1

By calculating the moles of urea

Moles of 1st urea $=1.29×\frac{345}{1000}=0.45moles$

Moles of 2nd urea $=0.653×\frac{485}{1000}=0.31moles$

Moles of 3rd urea $=0.775×\frac{835}{1000}=0.65moles$