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99 Percentile Qs Bank for JEE MainChemistrySome Basic Concepts of Chemistry

0.14   g r a m of N a H C O 3 tablet is added to neutralise 5   m l of 0.1  N   H C l produced in stomach of a man. Then percentage purity of N a H C O 3 tablet is -

Options

  1. A10 %
  2. B20 %
  3. C30 %
  4. D40 %

Correct answer

C. 30 %

Step-by-step solution

N a H C O 3 + H C l   ⟶ N a C l + H 2 O + C O 2 So moles of N a H C O 3 P u r e u s e d = m o l e s   o f H C l = 0.1   ×   5   ×   10 – 3 So weight of pure N a H C O 3   u s e d = 0.1   ×   5   × 10 – 3   ×   84   g r a m s = 0.042   g r a m s %Purity= M a s s   o f   p u r e     s u b s tan c e T o t a l   m a s s   o f   s a m p l e %   P u r i t y   o f &

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