$19.85 m L$ of $0.1 N N a O H$ reacts with $20 m L$ of $H C l$ solution for complete neutralisation. The molarity of $H C l$ solution is :

9.9

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0.99

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0.099

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0.0099

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Solution

The correct option is B $0.099$
Molarity of base $= \frac{n o r m a l i t y}{A c i d i t y} = \frac{0.1}{1} = 0.1$

$M_{1} V_{1} = M_{2} V_{2}$

$0.1 \times 19.85 = M_{2} \times 20$

$M_{2} = 0.09925 \approx 0.099$

Hence, option C is correct.

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19.85 mL of 0.1 N NaOH reacts with 20 mL of HCl solution for complete neutralisation. The molarity of HCl solution is :

The correct option is B 0.099Molarity of base =normalityAcidity=0.11=0.1 M1V1=M2V20.1×19.85=M2×20 M2=0.09925≈0.099Hence, option C is correct.

$19.85 m L$ of $0.1 N N a O H$ reacts with $20 m L$ of $H C l$ solution for complete neutralisation. The molarity of $H C l$ solution is :

The correct option is B $0.099$
Molarity of base $= \frac{n o r m a l i t y}{A c i d i t y} = \frac{0.1}{1} = 0.1$

$M_{1} V_{1} = M_{2} V_{2}$

$0.1 \times 19.85 = M_{2} \times 20$

$M_{2} = 0.09925 \approx 0.099$

Hence, option C is correct.