Suggested solution for 2017 SH 2 Prelim Practical
1
(a)
(ii)
Table 1:
Weighings of FA1
Mass of weighing bottle and
FA1
/ g
3.696
Mass of weighing bottle and residual
FA1
/ g
3.497
Mass of
FA1
transferred / g
0.199
Table 2: Titration of FA3 with FA4 using screened methyl orange as
indicator
1
2
Final burette reading / cm
3
30.00
36.90
Initial burette reading / cm
3
9.00
16.00
Volume of
FA4
used / cm
3
21.00
20.90
(iii)
average volume =
21.00 + 20.90
2
= 20.95 cm
3
(b)
(i)
amount of HNO
3
in 25.0 cm
3
= amount of NaOH = 20.95 / 1000 x
(9.10/40.0) = 4.77 x 10
‒
3
amount of HNO
3
in 250.0 cm
3
= 4.77 x 10
‒
2
(ii)
amount of HNO
3
in 25.0 cm
3
= 25/1000 x 2 = 0.05 mol
(iii)
amount of HNO
3
= 0.05 ‒ 4.77 x 10
‒
2
= 2.34 x 10
‒
3
(iv)
amount of MgCO
3
.xH
2
O = 2.34 x 10
‒
3
/ 2 = 1.17 x 10
‒
3
M
r
= 0.208 / 1.15 x 10
‒
3
= 170.2
x
= [180.7
–
(24.3 + 12.0 + 16.0 x 3)] / 18 = 4.77
≈ 5
(c)
The acid
will not be diluted
and close to 2.00 mol dm
−
3
, the titre value required
to neutralise 25 cm
3
of FA 3 will be
about 250 cm
3
/ more than 10 times greater
.
Since this
exceeds the capacity of burette
, it is unsuitable.

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