SATHEE: Solutions Question 144

Solutions Question 144

Question: How much of NaOH is required to neutralise 1500 $c m^{3}$ of 0.1 N HCl (At. wt. of Na =23) [KCET 2001]

Options:

A) 4 g

B) 6 g

C) 40 g

D) 60 g

Show Answer

Answer:

Correct Answer: B

Solution:

1500 $c m^{3}$ of 0.1 N HCl have number of gm equivalence $= \frac{N_{1} \times V_{1}}{1000} = \frac{1500 \times 0.1}{1000} = 0.15$
$∴ 0.15, g m .$ equivalent of NaOH $= 0.15 \times 40 = 6, g m .$

Solutions Question 144

Question: How much of NaOH is required to neutralise 1500 $c m^{3}$ of 0.1 N HCl (At. wt. of Na =23) [KCET 2001]

Options:

Answer:

Solution:

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Solutions Question 144

Question: How much of NaOH is required to neutralise 1500 cm3 of 0.1 N HCl (At. wt. of Na =23) [KCET 2001]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: How much of NaOH is required to neutralise 1500 $c m^{3}$ of 0.1 N HCl (At. wt. of Na =23) [KCET 2001]