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 $ .



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