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JEE Main 202324 JanuaryMorning Shift
Exam
JEE Main
Year
2023
Date
24 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2023 · 24 January · Morning Shift · Chemistry · Solutions

5gofNaOH was dissolved in deionized water to prepare a 450mL stock...

5gofNaOH5gofNaOH was dissolved in deionized water to prepare a 450mL450mL stock solution. What volume (in mL) of this solution would be required to prepare 500mL500mL of 0.1M0 . 1M solution ? Given : Molar Mass of Na,OandHis23,16and1gmol–1Na, OandHis23, 16and1g{mol}^{– 1} respectively

Correct answer

0

Step-by-step solution

Molarity od stock solution =molesofsoluteVolumeofsolutioninL=5/400.45= \dfrac{molesofsolute}{VolumeofsolutioninL} = \dfrac{5 / 40}{0 . 45}

M1V1=M2V2{M}_{1}{V}_{1} = {M}_{2}{V}_{2}

5/400.45×V1=0.1×5001000\dfrac{5 / 40}{0 . 45} \times {V}_{1} = 0 . 1 \times \dfrac{500}{1000}

V1=1801000=0.18L{V}_{1} = \dfrac{180}{1000} = 0 . 18L

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