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NEET 2008
Exam
NEET
Year
2008
Subject
Chemistry
Chapter
Coordination Compounds

NEET 2008 · Chemistry · Coordination Compounds

Volume occupied by one molecule of water (density =1 ~g ~cm^ -3 ) is

Volume occupied by one molecule of water (density =1 g cm−3=1 \mathrm{~g} \mathrm{~cm}^{-3} ) is

Options

  1. A9.0×10−23 cm39.0 \times 10^{-23} \mathrm{~cm}^3
  2. B6.023×10−23 cm36.023 \times 10^{-23} \mathrm{~cm}^3
  3. C3.0×10−23 cm33.0 \times 10^{-23} \mathrm{~cm}^3
  4. D5.5×10−23 cm35.5 \times 10^{-23} \mathrm{~cm}^3

Correct answer

C. 3.0×10−23 cm33.0 \times 10^{-23} \mathrm{~cm}^3

Step-by-step solution

6.02×10236.02 \times 10^{23} molecules of water =1 mol=18 g\begin{aligned} & =1 \mathrm{~mol} \\ & =18 \mathrm{~g} \end{aligned}\thereforeMassofonemoleculeofwaterMass of one molecule of water=186.023×1023 g∵V=md=d=mV6.023×1023×1∴≈3×10−23 cm3\begin{aligned} & =\frac{18}{6.023 \times 10^{23} \mathrm{~g}} \\ \because V=\frac{m}{d} & =\frac{d=\frac{m}{V}}{6.023 \times 10^{23} \times 1} \\ \therefore & \approx 3 \times 10^{-23} \mathrm{~cm}^3 \end{aligned}$
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