100 cm 3 of a 0-1 M HCl solution is mixed with 100 cm 3 of a 0-2 N N aOH solution- The concentration of excess reagent left and the nature of resulting solution is-A- Molarity -0-1 M and the solution is basicB- Molarity -0-1 M and the solution is acidicC- Molarity -0-05 M and the solution is basicD- Molarity -0-05 M and the solution is acidic

The correct option is C Molarity = 0.05 M and the solution is basicMolarity = number of moles of solutevolume of solution in mL×1000 Number of moles of HCl=1 ...

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Byju's Answer

Standard XII

Chemistry

Molarity

100 cm 3 of a...

Question

$100 {cm}^{3}$ of a 0.1 M $H C l$ solution is mixed with $100 {cm}^{3}$ of a 0.2 N $N a O H$ solution. The concentration of excess reagent left and the nature of resulting solution is:

Molarity = 0.1 M and the solution is basic

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Molarity = 0.1 M and the solution is acidic

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Molarity = 0.05 M and the solution is basic

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Molarity = 0.05 M and the solution is acidic

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Solution

The correct option is C Molarity = 0.05 M and the solution is basic
$Molarity = \frac{number of moles of solute}{volume of solution in mL} \times 1000$

$Number of moles of H C l = \frac{100 \times 0.1}{1000} = 0.01$

$Number of moles of N a O H = \frac{100 \times 0.2}{1000} = 0.02$

Since the moles of base are greater, the solution will be basic.
Moles of NaOH left after neutralization
$= 0.02 - 0.01 = 0.01$
Molarity of NaOH in the resultant solution
$= \frac{0.01}{200} \times 1000 = 0.05 M$

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100 cm3 of a 0.1 M HCl solution is mixed with 100 cm3 of a 0.2 N NaOH solution. The concentration of excess reagent left and the nature of resulting solution is:

$100 {cm}^{3}$ of a 0.1 M $H C l$ solution is mixed with $100 {cm}^{3}$ of a 0.2 N $N a O H$ solution. The concentration of excess reagent left and the nature of resulting solution is: