Question

the common factor of a²m⁴and a⁴m² is​

Answer 1

Step-by-step explanation:

0.2 moles of solute (urea) =0.2 X molar mass of urea = 0.2 X 60=12 grams

0.8 moles of solvent (water) = 0.8 X molar mass of water = 0.8 X 18= 14.4 grams.

Mass of solution = Mass of solute + Mass of solvent

=12 + 14.4

26.4 grams

Volume =mass/density

=26.4/d mL =26.4/1000 dL

Molarity =0.1/26.4/1000 d

1/10 X 1000d/26.4

Molarity =100d/26.4

If density=0.21gm/mL putting this value in the above equation we get.

Molarity=100X0.21/26.4=0.79 M

if d=0.21 gm/mL Its volume=26.4/1000 X0.21

volume=0.12 L =0.12 X1000= 120 ml

Percentage of solution= mass/volume of solution

=26.4/120

=0.22%