Quantitative Part 2 Flashcards
The concentration of blood glucose is usually given in millimoles per dm^3. A reading of 5.0 mmol dm^-3 is within the normal range. Glucose has a molar mass of 180g mol^-1.
What mass of glucose dissolved in 1 dm^3 of blood would give this normal reading? A) 0.090g B) 0.18g C) 0.90g D) 9.0g
C) 0.90g ✔️
Calculate the mass of Ca(OH)2 present in 100cm^3 of a 0.100 mol dm^-3 solution.
Assume the molar mass of Ca(OH)2 is 74.0 gmol^-1.
M = V/1000 x n x Mr
= 100/1000 x 0.1 x 74
= 0.740g ✔️
NaHSO4+NaOH ->Na2SO4 +H2O
0.0100 mol of NaHSO4 is neutralised with dilute NaOH, conc. of 0.200 moldm^-3.
Calculate the volume of NaOH required.
M = C x V V = 1000C/M (Convert for dm3)
V = (1000 x 0.01)/0.2
= 50.0 cm^3 ✔️
How many ions are present in 20cm^3 of 0.0500 moldm^-3 calcium chloride solution, CaCl2 ?
N = C x V/1000 (convert for dm3)
= 0.05 x 20/1000
= 0.001
However, we’re talking about IONS, and there are 3 ions in CaCl2;
= 0.001 x 3 = 0.003 ✔️
During a titration, when the solution in a pipette is transferred to a conical flask, a small amount of liquid remains in the tip of the pipette.
This situation should be dealt with by?
Leaving the liquid in the pipette which is calibrated to allow for it! ✔️
How do you calculate % error?
2 equations
(Accepted value - experimental value)
————————————— x 100
Accepted value
•••OR
% errors times by each other
—————————————-
Volume of vessel
(Ans/no. % errors) x 100
The tolerance of a 25cm^3 pipette is +/- 0.06cm^3. The percentage error in the measurement of 25 cm^3 using this pipette is?
0.06/25 = (2.4x10^-3) x 100
= +/- 0.24% ✔️
Explain what is meant by the term salt.
When a H+ ion in an acid is replaced by a metal ion. ✔️
What is a base?
Accepts the H+ ion ✔️
Calculate the relative formula mass of (NH4)2SO4.
2(14)+8(1)+32.1+4(16)= 132.1 gmol^-1 ✔️✔️
A student carries out a titration to find the concentration of come H2SO4. The student finds that 25.00cm^3 of a 0.088 mol dm^-3 NaOH is neutralised by 17.5 cm^3 of dilute H2SO4.
H2SO4+2NaOH->Na2SO4+2H2O
A) calculate the amount (in mol) of NaOH used.
B) determine the amount (in mol) of H2SO4 used.
C) calculate the concentration (in mol dm^-3) of the sulphuric acid.
A) (25/1000) x 0.088 = 0.0022mol
⬆️CONVERT (cm -> dm) ✔️
B) 1:2 ratio, (2.2x10^-3)/2 = 1.1x10^-3 ✔️
C) C = M/V
= 0.0011/0.0176 = 6.25x10^-2 ✔️
⬆️ rember to convert the 17.6
After a titration, a student left the resulting solution to crystallise. White crystals were formed, with the formula Na2SO4•xH2O and a molar mass of 322.1gmol^-1.
A) what term is given to the “•xH2O”
B) using the molar mass of the crystals, calculate the value of x.
A) water of crystallisation ✔️
B) Na2SO4 • H2O Mr 142.1 18 Find the molar mass of the xH2O: 322.1 - 142.1 = 180 Divide 180 by the mr of singular H2O: 180/18 = 10 Therefore x = 10 ✔️
State the formulae of 2 ions released when sulphuric acid is in aqueous solution.
H2SO4(aq) —> H+(aq) + (SO4)2-(aq)
✔️
A student adds a sample of solid potassium carbonate (K2CO3) to an excess of dilute sulphuric acid.
Describe what the student would see and write the equation for the reaction which takes place.
K2CO3 + H2SO4 —> K2SO4 + CO2 + H2O
The solid potassium carbonate would dissolve
✔️
In a titration 25cm^3 of 0.05 moldm^-3 tartaric acid, HxA exatly reacts with 12.5cm^3 of 0.2 moldm^-3 sodium hydroxide, NaOH. A solution of sodium tartate is produced.
A) calculate the amount in mol of HxA used
B) calculate the amount in mol of NaOH used
C) deduce the value for x in the formula of tartaric acid HxA.
A) m = c x V
0.025 x 0.05 = 1.25x10^-3 mol ✔️
•Remember conversions (cm->dm)
B) 0.0125 x 0.2 = 2.5x10^-3 mol ✔️
•Remember conversions (cm->dm)
C) using the mols from the last 2 answers:
1.25 : 2.5
Divide by smallest: 1 : 2
Therefore x = 2 ✔️
Antimony can be found in Sibnite. Sibnite contains 5% antimony.
Antimony can be obtained by reducing Sb2S3 with scrap iron
Sb2S3 + 3Fe —> 2Sb + 3 FeS
A) how many moles of Sb2S3 are in 500kg of a typical sample of Sibnite?
B) calculate the mass of antimony that could be obtained by processing 500kg of Sibnite.
A) n = mass/mr
Mr: (121.8x2) + (32.1x3) = 339.9
Mass: 500 x 0.05 =25kg (25000g)
n: 25000/339.9 = 73.6 mol ✔️
B) 1:2 ratio
73.6 x 2 = 147.2
121.8 x 147.2 = 17928.96g
= 17.9 kg ✔️
