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Some Basic Concepts of Chemistry question

2023 · 24 Jan · Shift 1 · Q16
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Some Basic Concepts of Chemistry question

2023 · 24 Jan · Shift 1 · Q16

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution? ‾\underline{\hspace{2cm}}​ Given : Molar Mass of Na, O and H is 23, 16 and 1 g mol −1^{-1}−1 respectively
Numerical answer
View written solutionFree

Correct answer: 180

  1. Find molar mass of NaOH

M(NaOH)=23+16+1=40 g mol−1M(\text{NaOH}) = 23 + 16 + 1 = 40\ \text{g mol}^{-1}M(NaOH)=23+16+1=40 g mol−1

  1. Calculate moles of NaOH in stock solution

Given mass of NaOH = 5 5\,5g,

n=540=0.125 moln = \frac{5}{40} = 0.125\ \text{mol}n=405​=0.125 mol

  1. Find molarity of the stock solution

Volume of stock solution = 450 450\,450mL =0.450 = 0.450\,=0.450L

M1=0.1250.450=0.2777‾ MM_1 = \frac{0.125}{0.450} = 0.277\overline{7}\ \text{M}M1​=0.4500.125​=0.2777 M

  1. Use dilution formula

We need to prepare 500 500\,500mL of 0.1 0.1\,0.1M solution.

Using

M1V1=M2V2M_1V_1 = M_2V_2M1​V1​=M2​V2​

where

  • M1=0.2777‾ M_1 = 0.277\overline{7}\,M1​=0.2777M
  • M2=0.1 M_2 = 0.1\,M2​=0.1M
  • V2=500 V_2 = 500\,V2​=500mL

So,

V1=M2V2M1=0.1×5000.2777‾V_1 = \frac{M_2V_2}{M_1} = \frac{0.1 \times 500}{0.277\overline{7}}V1​=M1​M2​V2​​=0.27770.1×500​

V1=500.2777‾=180 mLV_1 = \frac{50}{0.277\overline{7}} = 180\ \text{mL}V1​=0.277750​=180 mL

  1. Final Answer

The volume of stock solution required is:

180\boxed{180}180​

This matches the stored correct answer.

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