The volume of one mole of water at \(277 \mathrm{~K}\) is \(18 \mathrm{ml}\). One \(\mathrm{ml}\) of water contains 20 drops. The number of molecules in one drop of water will be \(\left(N_{\mathrm{A}}=6 \times 10^{23}\right)\) (a) \(1.07 \times 10^{21}\) (b) \(1.67 \times 10^{21}\) (c) \(2.67 \times 10^{21}\) (d) \(1.67 \times 10^{20}\)

Short Answer

Expert verified
(a) \(1.67 \times 10^{22}\) (b) \(3.33 \times 10^{22}\) (c) \(6.67 \times 10^{22}\) (d) \(1.33 \times 10^{23}\) Answer: (a) \(1.67 \times 10^{22}\)

Step by step solution

01

Find the number of moles in 1 ml of water

We know that one mole of water occupies a volume of \(18 \mathrm{ml}\) at \(277 \mathrm{K}\). To find the number of moles in \(1 \mathrm{ml}\) of water, we can use the ratio of volumes: $$ \text{Number of moles in 1 ml} = \frac{1 \mathrm{_~ml}}{18 \mathrm{~ml}} $$
02

Calculate the number of molecules in 1 ml of water

Now that we have the number of moles in \(1 \mathrm{ml}\) of water, we can find the total number of molecules present in \(1 \mathrm{ml}\) using Avogadro's number: $$ \text{Number of molecules in 1 ~ml} = \text{Number of moles in 1 ~ml} \times N_{\mathrm{A}} $$
03

Find the number of molecules in 1 drop of water

We know that there are 20 drops in \(1 \mathrm{ml}\) of water. To find the number of molecules in one drop, we can divide the total number of molecules in \(1 \mathrm{ml}\) by the number of drops: $$ \text{Number of molecules in 1 drop} = \frac{\text{Number of molecules in 1 ~ml}}{20} $$
04

Calculate and choose the correct answer

Putting all the values, we get: $$ \text{Number of molecules in 1 drop} = \frac{1}{18} \times N_{\mathrm{A}} \times \frac{1}{20} = \frac{1}{18 \times 20} \times 6 \times 10^{23} $$ Now, calculate this value and choose the option that matches the calculated result.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A volume of \(200 \mathrm{ml}\) of oxygen is added to \(100 \mathrm{ml}\) of a mixture containing \(\mathrm{CS}_{2}\) vapour and \(\mathrm{CO}\), and the total mixture is burnt. After combustion, the volume of the entire mixture is \(245 \mathrm{ml}\). Calculate the volume of the oxygen that remains (a) \(67.5 \mathrm{ml}\) (b) \(125.0 \mathrm{ml}\) (c) \(200.0 \mathrm{ml}\) (d) \(100.0 \mathrm{ml}\)

Two successive reactions, \(\mathrm{A} \rightarrow \mathrm{B}\) and \(\mathrm{B} \rightarrow \mathrm{C}\), have yields of \(90 \%\) and \(80 \%\) respectively. What is the overall percentage yield for conversion of \(\mathrm{A}\) to \(\mathrm{C}\) ? (a) \(90 \%\) (b) \(80 \%\) (c) \(72 \%\) (d) \(85 \%\)

The explosion of a mixture consisting of one volume of a gas being studied and one volume of \(\mathrm{H}_{2}\) yielded one volume water vapour and one volume of \(\mathrm{N}_{2}\). The formula of gas being studied, is (a) NO (b) \(\mathrm{NO}_{2}\) (c) \(\mathrm{N}_{2} \mathrm{O}\) (d) \(\mathrm{N}_{2} \mathrm{O}_{3}\)

A quantity of \(0.25 \mathrm{~g}\) of a substance when vaporized displaced \(50 \mathrm{~cm}^{3}\) of air at \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm} .\) The gram molecular mass of the substance will be (a) \(50 \mathrm{~g}\) (b) \(100 \mathrm{~g}\) (c) \(112 \mathrm{~g}\) (d) \(127.5 \mathrm{~g}\)

The vapour density of a sample of \(\mathrm{SO}_{3}\) gas is 28 . Its degree of dissociation in to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) is (a) \(1 / 7\) (b) \(1 / 6\) (c) \(6 / 7\) (d) \(2 / 5\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free