A sample of CaCO3 has Ca = 40%, C = 12% and O=48%. If the law of constant proportions is true, then the weight of Calcium in 4 g of a sample of calcium carbonate from another source will be: (a) 0.016 g (b) 0.16 g (c) 1.6 g (d) 16 g

Given

Ca = 40%, C = 12% and O = 48%

The law of constant proportions holds true

In 100 g CaCO3, 40 g Ca is present

In 4 g CaCO3, let x g Ca is present

⇒ x/4 = 40/100

x = (40 x 4)/100

x = 16/10

We get,

x = 1.6 g

Therefore, the correct option is (c) 

Was this answer helpful?

 
   

2.5 (4)

(7)
(3)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published.

*

*

close
close

DOWNLOAD

App Now

Ask
Question