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JEE Main 201912 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 12 January
Shift: Morning Shift
Subject: Chemistry
Chapter: Redox Reactions

JEE Main 2019 · 12 January · Morning Shift · Chemistry · Redox Reactions

50mL of 0 .5 M oxalic acid is needed to neutralize 25mL of sodium...

50mL

of

0 .5 M

oxalic acid is needed to neutralize

25mL

of sodium hydroxide solution. What is the amount of

NaOH

in

50mL

of the given sodium hydroxide solution?

Options

A $2g$
B $\text{4}g$
C $1g$
D $8g$

Correct answer

B. $\text{4}g$

Step-by-step solution

millimoles of oxalic acid

= 50 \times 0.5

milli equivalent of oxalic acid

= 50 \times 0.5 \times 2

millimoles of

NaOH = 25 \times M =

milli equivalent of

NaOH

∴ 50 \times 0 .5 \times 2 = 25 \times M

⇒ M = 2

M = 2

so,

1000\text{ml}

NaOH

solution contains

= 2\text{mol of}NaOH

= 2 \times 40 = 80\text{g}NaOH

∴ 50\text{ml}NaOH

solution contains

= \dfrac{80}{1000} \times 50 = 4\text{g}

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