Answer:
Example
0.5 mol of sodium hydroxide is dissolved in 2 dm3 of water. Calculate the concentration of the sodium hydroxide solution formed.
Concentration =
Concentration = 0.25 mol/dm3
Volume units
Volumes used in concentration calculations must be in dm3, not in cm3. It is useful to know that 1 dm3 = 1000 cm3. This means:
divide by 1000 to convert from cm3 to dm3
multiply by 1000 to convert from dm3 to cm3
For example, 250 cm3 is 0.25 dm3 (250 ÷ 1000). It is often easiest to convert from cm3 to dm3 before continuing with a concentration calculation.
Question
100 cm3 of dilute hydrochloric acid contains 0.02 mol of dissolved hydrogen chloride. Calculate the concentration of the acid in mol/dm3.
Reveal answer
Converting between units
The relative formula mass of the solute is used to convert between mol/dm3 and g/dm3:
to convert from mol/dm3 to g/dm3, multiply by the relative formula mass
to convert from g/dm3 to mol/dm3, divide by the relative formula mass
Remember: the molar mass is the Ar or Mr in grams per mol.
Example
Calculate the concentration of 0.1 mol/dm3 sodium hydroxide solution in g/dm3. (Mr of NaOH = 40)
Concentration = 0.1 × 40
= 4 g/dm3
Answer:
0.681 atm
Explanation:
To solve this problem, we make use of the General gas equation.
Given:
P1 = 785 torr
V1 = 2L
T1 = 37= 37 + 273.15 = 310.15K
P2 = ?
V2 = 3.24L
T2 = 58 = 58+273.15 = 331.15K
P1V1/T1 = P2V2/T2
Now, making P2 the subject of the formula,
P2 = P1V1T2/T1V2
P2 = [785 * 2 * 331.15]/[310.15 * 3.24]
P2 = 515.715 Torr
We convert this to atm: 1 torr = 0.00132 atm
515.715 Torr = 515.715 * 0.00132 = 0.681 atm
Answer:
Solution - (a) Brine . Suspension - (c) sand and water, (g) chalk and water Colloid - (e) air, (f) smoke , (d) soda , (b) milk
Explanation:
A solution is an homogeneous mixture of two or more compounds.
A suspension is a heterogeneous mixture of two or more compounds while a colloid is a homogeneous mixture of two or more compounds with suspended particles which do not settle.
So, under these definitions, the classifications are as follows-
Solution - (a) Brine .
Suspension - (c) sand and water, (g) chalk and water
Colloid - (e) air, (f) smoke , (d) soda , (b) milk
