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JEE Main 202427 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
27 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Main 2024 · 27 January · Evening Shift · Chemistry · Some Basic Concepts of Chemistry

Volume of 3MNaOH (formula weight 40g mol ^ - 1 ) which can be...

Volume of 3MNaOH3MNaOH (formula weight 40gmol−140g{mol}^{- 1} ) which can be prepared from 84g84g of NaOHNaOH is ____×10−1dm3\times {10}^{- 1}{dm}^{3}.

Correct answer

0

Step-by-step solution

Number of moles = Mass / Molar mass

Molar mass of NaOH=4040 g/mol

By Using the definition of molarity to find the volume of the solution:

Molarity= Number of moles of soluteVolume of solution (in liters)\dfrac{\text{Number of moles of solute}}{\text{Volume of solution (in liters)}}

M=nNaOHVsol(inL)⇒3=(84/40)V⇒V=0.7L=7×10−1LM = \dfrac{{n}_{NaOH}}{{V}_{sol} ( \text{in}L )} ⇒ 3 = \dfrac{( 84 / 40 )}{V} ⇒ V = 0 . 7L = 7 \times {10}^{- 1}L

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