Solutions Question 228
Question: The weight of sodium carbonate required to prepare 500 ml of a semi- normal solution is [JIPMER 1999]
Options:
A) 13.25 g
B) 26.5 g
C) 53 g
D) 6.125 g
Show Answer
Answer:
Correct Answer: A
Solution:
$ N=\frac{w\times 1000}{eq.,wt.\times volumeinml.} $
$ eq.,wt.=\frac{106}{2}=53 $
$ w=\frac{0.5\times 53\times 500}{1000}=13.25 $ .
