LIVE Course for free

Rated by 1 million+ students
Get app now
0 votes
1.1k views
in Solutions by (47.9k points)
closed by

Calculate (he molarity of each of the following solutions:

1. 30 g of CO(NO3)2 . 6H2O = in 4.3 L of solution 

2. 30 mL of 0.5 M H2SO4 diluted to 500 mL.

1 Answer

+1 vote
by (46.8k points)
selected by
 
Best answer

1. Molar mass of CO(NO3)2 . 6H2O = 58.7 + 2(14 + 48) + (6 x 8)g mol-1 

= 58.7 + 124 + 108 g mol-1 = 290.7 g mol-1

Volume of solution = 4.3 L

2. 1000 mL of 0.5 M H2SO4 Contain H2SO4 = 0.5 moles 

30 mL of 0.5 M H2SO4 contain H2SO4\(\frac{0.5}{100}\) x 30 mole 

= 0.015 mole

Volume of solution = 500 mL = 0.500 L

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...