Question

2.50 gram of sample of Sodium Bicarbonate when strongly heated with 300 cm cube of seo2 measured at 27 degree Celsius and 760 mm pressure calculate the percentage purity of the sample​

Answer 1

Answer:

We can write the equation as 2NaHCO3 -> Na2CO3 + CO2 +H2O .

The molecular mass of NahCO3 will be 84*2 = 168g.

Thus we know that 22400 ml CO2 is given by - 168 g of NaHCO3.

Hence, we can say that 300 ml CO2 is given by - 168 x 300 /22400 = 2.25g.

Hence, the Percentage purity of the sample will be = 2.25 x 100 /2.5 .

On calculating we will get the purity percentage as 90%.

Answer 2

Answer:

Ans is 81.9 %

Explanation:

2NaHCO3 → Na2CO3 + CO2 + H2O

168 g 22400 ml

Let the volume of CO2 at STP be V ml

∴

P

1

V

1

T

1

=

P

2

V

2

T

2

or,

760

×

V

273

=

760

×

300

(

273

+

27

)

or,

760

×

V

273

=

760

×

300

300

or, V = 273 ml

Now, 22400 ml of CO2 is produced from 168 g of NaHCO3

∴ 273 ml of CO2 is produced from

168

×

273

22400

g = 2.047 g of NaHCO3

∴ Purity of sodium bicarbonate =

2.047

2.5

×

100

=

81.9

%