A sample of calcium carbonate (CaCO3) has the following percentage composition : Ca = 40%, C : 12%, O = 48%. If the law of constant proportions is true, then the weight of calcium in 4 g of a sample of calcium carbonate obtained from another source will be :

A sample of calcium carbonate (CaCO₃) has the following percentage composition : Ca = 40%, C : 12%, O = 48%. If the law of constant proportions is true, then the weight of calcium in 4 g of a sample of calcium carbonate obtained from another source will be :

A
0.016 g
B
0.16 g
C
1.6 g
D
16 g

Solution:

Molecular weight of CaCO₃ = 100 gm

100 gm of sample contains = 40 gm of Ca

4 gm of sample contains = ?

= $\frac{4 \times 40}{100} = 1.6$ gm

Related content

Top Online Courses for Class 11 JEE Preparation in Chennai

Success Stories of IIT Toppers: Inspiration for Aspiring Students

Infinity Learn: Best Online NEET Coaching for Class 11–12 Droppers in South India

Top 5 Reasons Why Online Coaching is Better for JEE Main and JEE Advanced

MC Full Form – Meaning, Uses, and Domain-Specific Definitions

NADP Full Form – Nicotinamide Adenine Dinucleotide Phosphate

IIT Full Form – Indian Institute of Technology

OBC Full Form – Other Backward Classes

ISO Full Form – International Standardization Organization

MS Full Form – Master of Science