Answer to Question #127550 in Chemistry for Sara

Question

3g of Aluminium nitrate having molecular weight 212.996 g/mol was dissolved in water and total volume is 50g. If we take 3.5ml of that solution and dilute it up to 10ml while the density of new solution is 0.96g/cm3 what will be w/w and w/v of the new solution

Answer 1

Concentration of Al(NO₃)₃ in the initial solution is:

w/v(Al(NO₃)₃)₁=3/50=0.06 g/ml

Mass of Al(NO₃)₃ in the sample:

m (Al(NO₃)₃) = 0.06 х 3.5 = 0.21 g

Concentration (w/v) of Al(NO₃)₃ in the new solution:

w/v (Al(NO₃)₃)₂ = 0.21/10 = 0.021 g/ml

Concentration (w/w) of Al(NO₃)₃ in the new solution:

w/w (Al(NO₃)₃)₂ = 0.21/(10x0.96) = 0.022 g/g