Concentration and chemistry

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Marks Available:	7
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Concentration and chemistry

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Question 1

a)

A 1000 mL solution is made by mixing 40 g of $\mathrm{MgCl_2}$ and water. The percentage weight per volume of the solution is % w/v of $\mathrm{MgCl_2}$ .

Percentage weight per volume is the weight of the solute in grams divided by the amount of solution in millilitres expressed as a percentage. That is, we do the division and multiply by 100.

In particular, if a 1000 mL solution is made by mixing 40 g of $\mathrm{MgCl_2}$ and water. The percentage weight per volume of the solution is:

$\dfrac{\text{grams}}{\text{millilitres}}\times 100$	$=$	$\dfrac{\var{gramsnice}}{\var{millnice}}\times 100$ % w/v
	$=$	$\var{ansnice}$ % w/v

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b)

A 800 mL solution is made by mixing 68 g of $\text{KCl}$ and water. The percentage weight per volume of the solution is % w/v of $\text{KCl}$ .

Percentage weight per volume is the weight of the solute in grams divided by the amount of solution in millilitres expressed as a percentage. That is, we do the division and multiply by 100.

In particular, if a 800 mL solution is made by mixing 68 g of $\text{KCl}$ and water. The percentage weight per volume of the solution is:

$\dfrac{\text{grams}}{\text{millilitres}}\times 100$	$=$	$\dfrac{\var{gramsb}}{\var{millb}}\times 100$ % w/v
	$=$	$\var{ansb}$ % w/v

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c)

A 700 mL solution is made by mixing 18.3 g of $\mathrm{Na_2SO_4}$ and water. The percentage weight per volume of the solution is % w/v of $\mathrm{Na_2SO_4}$ (to 2 decimal places).

Percentage weight per volume is the weight of the solute in grams divided by the amount of solution in millilitres expressed as a percentage. That is, we do the division and multiply by 100.

In particular, if a 700 mL solution is made by mixing 18.3 g of $\mathrm{Na_2SO_4}$ and water. The percentage weight per volume of the solution is:

$\dfrac{\text{grams}}{\text{millilitres}}\times 100$	$=$	$\dfrac{\var{grams}}{\var{mill}}\times 100$ % w/v
	$=$	$\var{ans}$ % w/v (to 2 dec. pl)

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d)

A 1.2 % $\text{NaCl}$ solution contains 12 g of $\text{NaCl}$ in every...

It is important to know 1.2 % $\text{NaCl}$ solution contains 1.2 g of $\text{NaCl}$ in every 100 mL of solution.

Given this, since we have 12 g of $\text{NaCl}$ we have ten times the standard amount and so we must have ten times the solution, that is 1000 mL.

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100 mL.
200 mL.
500 mL.
1000 mL.

Expected answer:

100 mL.
200 mL.
500 mL.
1000 mL.

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Advice

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Question 2

1.4 moles of a solute are dissolved to create 4 L of solution. The molar concentration is Expected answer: mol/L (to one decimal place).

Molar concentration is the number of moles divided by the volume of solution in litres.

In particular, if a 4 L solution is made by mixing 1.4 moles of a solute then the molar concentration is calculated as follows:

$\dfrac{\text{number of moles}}{\text{litres}}$	$=$	$\dfrac{\var{number}}{\var{litres}}$ mol/L
	$=$	$\var{onedec}$ mol/L (1 dec. pl)

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Question 3

a)

The molecular weight of $\mathrm{C_6H_{12}O_6}$ is Expected answer: g/mol (round each atomic weight to the nearest whole number for this exercise).

The atomic mass of C is 12, the atomic mass of H is 1, the atomic mass of O is 16. Therefore, the molecular mass of $\mathrm{C_6H_{12}O_6}$ is $6\times 12+12\times 1+6\times 16=180$.

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b)

How many moles of $\mathrm{C_6H_{12}O_6}$ are present in 1710 g of $\mathrm{C_6H_{12}O_6}$?

Expected answer: moles

The number of moles is the weight of the compound in grams/mol divided by the molecular mass of the compound in grams.

In particular, the number of moles in 1710 g of $\mathrm{C_6H_{12}O_6}$ is calculated as follows:

$\dfrac{\text{mass of compound (g)}}{\text{molecular mass (g/mol)}}$	$=$	$\dfrac{\var{grams} \text{g}}{\var{chem[2]} \text{g/mol}}$
	$=$	$\var{mult}$ mol

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Concentration and chemistry

Concentration and chemistry

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b)

c)

d)

Advice

Advice

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b)

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