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NEET 2018
Exam
NEET
Year
2018
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

NEET 2018 · Chemistry · Some Basic Concepts of Chemistry

In which case the number of molecules of water maximum?

In which case the number of molecules of water maximum?

Options

  1. A0.00224L0 . 00224L of water vapours at 1atm1atm and 273K273K.
  2. B0.18g0 . 18g of water.
  3. C18mL18mL of water.
  4. D10−3{10}^{- 3} mol of water.

Correct answer

C. 18mL18mL of water.

Step-by-step solution

(i) 18mL18mL water:

As dH2O=1g/mL{d}_{{H}_{2}O} = 1g / mL So WH2O=18g{W}_{{H}_{2}O} = 18g

nH2O=1818=1{n}_{{H}_{2}O} = \dfrac{18}{18} = 1

Molecules =1×NA= 1 \times {N}_{A}

(ii) 0.18g0 . 18g of water:

nH2O=0.1818=0.01{n}_{{H}_{2}O} = \dfrac{0 .18}{18} = 0 .01

H2O{H}_{2}O molecules=0.01×NA\text{molecules} = 0 .01 \times {N}_{A}

(iii) (VH2O(g))STP=0.00224L{( {V}_{{H}_{2}O( g )} )}_{STP} = 0 .00224 L

nH2O=V22.4=0.0022422.4=0.0001{n}_{{H}_{2}O} = \dfrac{V}{22 .4} = \dfrac{0 .00224}{22 .4} = 0 .0001

Molecules =0.0001×NA= 0 .0001 \times {N}_{A}

(iv) nH2O=10−3{n}_{{H}_{2}O} = {10}^{- 3}

H2O{H}_{2}O molecules=10−3×NA\text{molecules} = {10}^{- 3} \times {N}_{A}
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