Question

If 2.75 g of urea CO(NH2)2 are dissolved to make 50.0 mL of solution..

what is the molarity of this solution ?

if you take 10.0mL of this solution how many moles of urea do you have?

if you dilute this 10.0mL portion to 200mL what is the molarity of the solution?

Answer 1

1)

Molar mass of CO(NH2)2,

MM = 1*MM(C) + 1*MM(O) + 2*MM(N) + 4*MM(H)

= 1*12.01 + 1*16.0 + 2*14.01 + 4*1.008

= 60.062 g/mol

mass(CO(NH2)2)= 2.75 g

use:

number of mol of CO(NH2)2,

n = mass of CO(NH2)2/molar mass of CO(NH2)2

=(2.75 g)/(60.062 g/mol)

= 4.579*10^-2 mol

volume , V = 50 mL

= 5*10^-2 L

use:

Molarity,

M = number of mol / volume in L

= 4.579*10^-2/5*10^-2

= 0.9157 M

Answer: 0.916 M

2)

V = 10.0 mL = 0.0100 L

Use:

number of mol = Molarity * Volume

= 0.9157 M * 0.0100 L

= 9.157*10^-3 mol

Answer: 9.16*10^-3 mol

3)

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M1 = 0.9157 M

V1 = 10.0 mL

V2 = 200 mL

use:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (0.9157*10)/200

M2 = 0.0458 M

Answer: 0.0458 M