Question

49 gm of H2SO4 are dissolved in enough water to make it 1 litre of solutions of density 1.049 g/cc.Find the molarity,normality,molality of H2SO4 in the solution

Answer 1

Answer :

Molarity = 0.499 mole/L

Molality = 0.476 mole/Kg

Normality = 0.998 eq/L

Solution : Given,

Mass of $H_{2}SO_{4}$ = 49 g

Volume of solution = 1 L

Density of solution = 1.049 g/cc

Molar mass of $H_{2}SO_{4}$ = 98.079 g/mole

Formula used :

A = Water      B = $H_{2}SO_{4}$

For Molarity :    $Molarity=\frac{W_{B} }{M_{B}\times V(in L)}$

where,    

$W_{B}$ = mass of $H_{2}SO_{4}$

$M_{B}$ = Molar mass of $H_{2}SO_{4}$

V = volume of solution (in L)

For Molality :     $Molality=\frac{W_{B} }{M_{B}\times W_{A} (in Kg)}$

where,

$W_{A}$ = Mass of water (in Kg)

For Normality :    Normality = Molarity × Number of $H^{+}$ ions (Basicity)

Calculation for Molarity :

put all the given values in above formula of Molarity, we get

$Molarity=\frac{49g}{98.079g/mole\times 1L}$ = 0.499 mole/L

Calculation for Molality :

Volume = 1 L = $10^{3}$ cc

Mass of water = $W_{A}$ = Density × Volume = 1.049 g/cc × $10^{3}$ cc = 1.049 × $10^{3}$ g = 1.049 Kg

Now put all the given values in above formula of Molality, we get

$Molality=\frac{49g}{98.079g/mole\times 1.049 (Kg)}$ = 0.476 mole/Kg

Calculation for Normality :

Number of $H^{+}$ ions (Basicity) in $H_{2}SO_{4}$ = 2

Now put all the values in above formula of Normality, we get

Normality = 0.499 × 2 = 0.998 eq/L