Question

in sc2 pg no. 67 of state board. 5 lesson acid base and salts
complete the following table there are two tables but i want the first table whole chart solution plzz give me the answer pzz​

Answer 1

Answer:

Write down the concentration of each of the following solutions in g/L and mol/L.

a. 7.3g HCl in 100ml solution

b. 2g NaOH in 50ml solution

c. 3g CH3COOH in 100ml solution

d. 4.9g H2SO4 in 200ml solution

ANSWER:

a)7.3 g HCl in 100 mL solution:

Concentration of HCl in g L−1 =Mass of solute in gramVolume of HCl in litre

=7.3 × 1000100=73 g L-1

In terms of gram per litre:

7.3 g of HCl in 100 mL has concentration = 73 g L-1

Molecular mass of HCl = 1+35.5=36.5

Molarity =Mass of solute in molesVolume of HCl in L

=7.3 × 100036.5 ×100=2 mol L-1

In terms of moles per litre:

7.3 g of HCl in 100 mL has concentration = 2 mol L-1

b)2g NaOH in 50 mL solution

Concentration of NaOH in g L−1 =Mass of NaOH in gVolume of NaOH in L

=2 × 100050= 40 g L-1

In terms of gram per litre:

2 g of NaOH in 50 mL has concentration = 40 g L-1

Molecular mass of NaOH = 23 +16 + 1 = 40

Molarity =Mass of solute in molesVolume of NaOH in L

=2 × 100040 ×50=1 mol L-1

In terms of moles per litre:

2 g of NaOH in 50 mL has concentration = 1 mol L-1

c)3 g CH3COOH in 50 mL solution

Concentration of CH3COOH in g L−1 =Mass ofCH3COOH in gVolume of CH3COOH in L

=3 × 1000100= 30 g L-1

In terms of gram per litre:

3 g of CH3COOH in 100 mL has concentration = 30 g L-1

Molecular mass of CH3COOH = 2 × 12 + 2 × 16 + 4 × 1 = 60

Molarity =Mass of solute in molesVolume of solution in L

=3 × 100060 ×100=0.5 mol L-1

In terms of moles per litre:

3 g of CH3COOH in 100 mL has concentration = 0.5 mol L-1

d)4.9 g H2SO4 in 200 mL solution