Question
\({\text{300 c}}{{\text{m}}^{\text{3}}}\) of water is added to a solution of \({\text{200 c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.5 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) sodium chloride. What is the concentration of sodium chloride in the new solution?
A. \({\text{0.05 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\)
B. \({\text{0.1 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\)
C. \({\text{0.2 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\)
D. \({\text{0.3 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\)
▶️Answer/Explanation
C
1cm3=.001 dm3.
\({\text{200 c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.5 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) sodium chloride contains : 0.5×200×.001 = 0.1 moles NaCl
Final volume of solution = 200 + 300 cm3= 500 cm3= 0.5 dm3.
concentration of sodium chloride in the new solution = 0.1 moles/0.5 dm3= \({\text{0.2 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\)
Question
Which solution neutralizes 50.0 cm3 of 0.120 mol dm–3 NaOH (aq)?
A. 12.5 cm3 of 0.080 mol dm–3 H3PO4
B. 25.0 cm3 of 0.120 mol dm–3 CH3COOH
C. 25.0 cm3 of 0.120 mol dm–3 H2SO4
D. 50.0 cm3 of 0.060 mol dm–3 HNO3
▶️Answer/Explanation
C
1 cm3 = 0.001 dm3
50.0 cm3 of 0.120 mol dm–3 NaOH = 0.120×0.050=.006 moles NaOH which contains .006 moles OH– ions.
It requires 0.006 moles of H+ ions for neutralization.
Now,
A gives : 3×0.080×0.0125=0.003 moles of H+ ions.
B gives : 1×0.120×0.025=0.003 moles of H+ ions.
C gives : 2×0.120×0.025=0.006 moles of H+ ions which is equal to .006 moles OH– ions. Hence correct answer is C.
D gives : 1×0.o60×0.050=0.003 moles of H+ ions.
Question
A compound with Mr = 102 contains 58.8 % carbon, 9.80 % hydrogen and 31.4 % oxygen by mass.
What is its molecular formula?
Ar: C = 12.0; H = 1.0; O = 16.0
A. C2H14O4
B. C3H4O4
C. C5H10O2
D. C6H14O
▶️Answer/Explanation
C
By weight C:H:O = 58.8:9.8:31.4
By moles C:H:O = 58.8/12 : 9.8/1 : 31.4/16
By moles C:H:O = 4.9 : 9.8 : 1.9625
By moles C:H = 1:2 = 2:4
By moles C:O = 1:0.4
Let x moles of C atoms are there.
Then, 2x moles of H and 0.4x moles of O.
Given, Molecular mass = 102 = 12x + 2x + 6.4x
20.4x = 102
x = 5
It has 5 C atoms, 10 H atoms and 2 O atoms i.e. C5H10O2