"Some Basic Concepts in Chemistry (Mole Concept)" — Practice

Question 1

How many atoms of oxygen are present in 1 mole of sulphuric acid, and how many molecules?

Accepted Answer

One mole of H$_2$SO$_4$ contains $6.022	imes10^{23}$ molecules. Each molecule has four oxygen atoms, so the oxygen count is $4 	imes 6.022	imes10^{23} = 2.409	imes10^{24}$ atoms. The question is testing whether you notice which entity is being asked for.

Question 2

Natural chlorine has average atomic mass 35.48 u and consists of chlorine-35 and chlorine-37. Find the percentage abundance of each isotope.

Accepted Answer

Let the fraction of chlorine-35 be $x$. Then $35x + 37(1-x) = 35.48$, giving $37 - 2x = 35.48$, so $x = 0.76$. The abundances are about 76 per cent chlorine-35 and 24 per cent chlorine-37, which matches the accepted values.

Question 3

A hydrocarbon contains 85.7 per cent carbon by mass and has molar mass 56 g mol$^{-1}$. Find its molecular formula.

Accepted Answer

Hydrogen is 14.3 per cent. Per 100 g: carbon $85.7/12.01 = 7.135$ mol, hydrogen $14.3/1.008 = 14.19$ mol. Dividing by the smaller gives 1 : 1.988, so the empirical formula is CH$_2$ with mass 14.03. Then $k = 56/14.03 = 4$, giving C$_4$H$_8$.

Question 4

Carbon forms two oxides. In the first, 12 g of carbon is combined with 16 g of oxygen; in the second, with 32 g. Which law does this illustrate and how?

Accepted Answer

The law of multiple proportions. For a fixed mass of carbon, the oxygen masses combining with it are 16 g and 32 g, in the simple whole-number ratio 1:2. The two oxides are carbon monoxide and carbon dioxide, and Dalton's atomic theory explained exactly this pattern.

Question 5

Explain why the atomic mass of chlorine in the periodic table is 35.5 when no chlorine atom has that mass.

Accepted Answer

Because the tabulated value is a weighted average over the naturally occurring isotopes, not the mass of any single atom. Chlorine-35 and chlorine-37 both exist in roughly a three to one ratio, and the average of 35 and 37 weighted by those abundances is 35.48.

Question 6

6.0 g of magnesium is burned in 8.0 g of oxygen gas. Which reagent is limiting, and what mass of magnesium oxide forms?

Accepted Answer

For $\mathrm{2Mg + O_2 ightarrow 2MgO}$: magnesium is $6.0/24.31 = 0.2468$ mol and oxygen is $8.0/32.00 = 0.250$ mol. Dividing by coefficients gives 0.1234 and 0.250, so magnesium is limiting. Magnesium oxide formed is 0.2468 mol, that is $0.2468 	imes 40.31 = 9.95$ g.

Question 7

In the reaction of 4.00 g of hydrogen with 32.0 g of oxygen to form water, find the mass of water produced and of any reagent left over.

Accepted Answer

For $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$: hydrogen is $4.00/2.016 = 1.984$ mol and oxygen is 1.000 mol. Dividing by coefficients gives 0.992 and 1.000, so hydrogen is limiting, just. Water formed is 1.984 mol, that is 35.7 g, and the oxygen left is $1.000 - 0.992 = 0.008$ mol, about 0.26 g.

Question 8

An aqueous solution is 20.0 per cent glucose by mass. Find its molality and the mole fraction of glucose.

Accepted Answer

Take 1000 g of solution: glucose is 200 g $= 200/180.16 = 1.110$ mol, and water is 800 g $= 0.800$ kg. Molality is $1.110/0.800 = 1.39$ mol kg$^{-1}$. Water is $800/18.02 = 44.39$ mol, so the mole fraction of glucose is $1.110/(1.110+44.39) = 0.0244$.

Question 9

Concentrated nitric acid is 68.0 per cent by mass with density 1.41 g mL$^{-1}$. Calculate its molarity.

Accepted Answer

Take 1 L of solution, of mass 1410 g. The nitric acid in it is $0.680 	imes 1410 = 958.8$ g. With $M(\mathrm{HNO_3}) = 63.01$, the amount is 15.2 mol, so the molarity is 15.2 mol L$^{-1}$. Taking 1 L as the basis is what makes this a two-line calculation.

Question 10

What volume of 0.500 M sodium hydroxide must be diluted to make 250 mL of 0.100 M solution?

Accepted Answer

Use $M_1V_1 = M_2V_2$, since dilution changes the volume but not the moles of solute. So $V_1 = (0.100)(250)/0.500 = 50.0$ mL. Take 50.0 mL of stock and add water to a final volume of 250 mL, rather than adding 200 mL of water.

Question 11

Heating 40.0 g of potassium chlorate gives 12.8 g of oxygen. Find the percentage yield.

Accepted Answer

For $\mathrm{2KClO_3 ightarrow 2KCl + 3O_2}$, with $M = 122.55$, the amount is $40.0/122.55 = 0.3264$ mol. Oxygen expected is $	frac{3}{2}$ of that, which is 0.4896 mol $= 15.67$ g. The percentage yield is $12.8/15.67 	imes 100 = 81.7$ per cent.

Question 12

Why do colligative properties use molality rather than molarity, given that molarity is far easier to measure?

Accepted Answer

Because molarity is defined per litre of solution and solution volume expands on heating, so the same solution has different molarities at different temperatures without any change in composition. Colligative properties depend on the ratio of solute to solvent particles, which does not change with temperature, so they must be expressed in a unit that also does not — molality, defined per kilogram of solvent.

Practice — "Some Basic Concepts in Chemistry (Mole Concept)"

Sign in to save attempts to your dashboard and track your accuracy over time

Find a verified tutor

Join a live course

Browse coaching centres

About

Categories

Support

Documentation

Get Study Tips & Updates

Download Our App

Download for iOS

Download for Android

Popular Cities