Active Mass

Active mass

source : chem4kids.com

Active Mass-

Active mass of a substance is its molar concentration or molarity. Number of moles present in  one litre of solution is active mass of that substance. Its unit is mole /litre. Active mass of a substance is represented by enclosing its formula in square brackets

Active Mass of Substance = number of moles /volume of solution in litre

No. of Moles = w/m =weight of solute /molecular weight

Active Mass = w /mV(l)

Q.1 In the pot of capacity 2 litre at 400⁰C, 4 gm of H₂ &  128 gm of HI was taken. Calculate the active mass or molar concentration ?

Ans.  w of H2 =4 gm.

V = 2 litre

m of H2 = 2

             [H2] =w/mV(l)

=4 /2 x2 =1 mole /litre Ans.

[HI] =w/mV

w of HI =128 gm.

m of HI =128

=128/128 x 2 =0.5 mole /litre Ans.

Q 2—In 500 ml of a Solⁿ 15 gm acetic Acid & 8 gm. Methanol is present (CH₃OH) what will be their active masses?

Ans V=500ml ==0.5 l

            w of CH₃COOH= 15gm

mol.wt. of CH₃COOH= 60

[CH₃COOH] = w/mV

=15 /60 x 0.5 =0.5 mole /litre Ans.

w of CH₃OH=8gm

m of CH₃OH=32

[CH₃OH] = w/mV

=8 /32 x 0.5 =0.5   mole /litre Ans.

Q 3—In 1 litre gaseous mix. 1 gm H₂ & 2.8 gm N₂ is present. Calculate the molar concentrations ?

Ans.                w of H2= 1gm

                        w of N2 = 2.8 gm

v= 1 litre

m of H₂=2

m of N2 =28

[H2] = w/mV

=1 /2 x 1 =0.5 mole /litre

[N2] = w/mV

=2.8 /28 x 1 =0.1 mole /litre

[H2] =0.5 mole/litre; [N2] =0.1  mole /litre

Q  4—4 litre of a gaseous mix. contains 4 gm. H₂, 142 gm. Cl₂ & 73 gm. HCl. Calculate their active masses?

V=4 litre

w of H₂ = 4 gm,   m of H₂ = 2

[H2] = w/mV

4 /2 x 4 =0.5 mole/litre

w of Cl₂ = 142 gm,   m of Cl₂ = 71

[Cl2] = w/mV

=142 /71 x 4 =0.5 mole /litre

w of HCl = 73 gm,   m of HCl = 36

[HCl] = w/mV

=73 /36.5 x 4 =0.5 mole /litre Ans.

Q 5—In the pot of capacity of 2 l, 64 gm. of HI is present. Calculate its active mass?

Ans. V=2 l,  w of HI = 64 gm.    m of HI= 128

[HI] = w/mV

=64 /128 x 2 =0.25 mole /litre

Q 6—In 500ml of solution 46  gm of C2H5OH & 15 gm. of Acetic Acid are present. Calculate their active masses.

Ans.                V =500 ml= 0.5 l

                        w of C₂H₅OH= 46

m of C₂H₅OH =46

[C2H5OH] = w/mV

=46 /46 x 0.5 =2 mole /litre

w of CH₃COOH= 15 gm

m of CH₃COOH=60

[CH3COOH] = w/mV

=15 /60 x 0.5 =0.5 mole /litre Ans.