CSL WS4.1 Solutions
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Worksheet 4.1: Solutions Mole calculations No. Answer 1 a N(electrons) = n L = 1.50 6.02 10 23 = 9.03 10 23 b N(atoms) = n L = 12.2 6.02 10 23 = 7.34 10 24 c N(atoms) = n L = 2.75 3 × 6.02 10 23 = 4.97 10 24 d N(ions) = n L = 0.034 × 2 6.02 10 23 = 4.09 10 22 2 a n(atoms) = = = 0.500 mol b n(molecules) = = = 2.0 × 10 –14 mol c n(ions) = = × 3 = 6.23 × 10 –4 mol d n(atoms) = = × 24 = 0.24 mol 3 a M r (C 3 H 8 ) = 3 × 12.01 + 8 × 1.01 = 44.11 b M r (CH 3 NH 2 ) = 12.01 + 5 × 1.01 + 14.01 = 31.07 c M r (SO 3 ) = 32.06 + 3 × 16.00 = 80.06 d M r (H 2 O 2 ) = 2 × 1.01 + 2 × 16.00 = 34.02 4 a 239.19 g b 134.45 g c 98.08 g d 17.04 g 5 a n(C) = = = 3.0 mol b n(O 3 ) = = = 0.20 mol c n(SO 3 ) = = = 7.7 10 –4 mol d n(V 2 O 5 ) = = = 16.6 mol Page 1 © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2008. This page from the Chemistry: For use with the IB Diploma Programme SL Teacher’s Resource may be reproduced for classroom use.
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CSL WS4.1 Solutions
Transcript of CSL WS4.1 Solutions
Worksheet 4.1: SolutionsMole calculations
No. Answer
1 a N(electrons) = n L = 1.50 6.02 1023 = 9.03 1023
b N(atoms) = n L = 12.2 6.02 1023 = 7.34 1024
c N(atoms) = n L = 2.75 3 × 6.02 1023 = 4.97 1024
d N(ions) = n L = 0.034 × 2 6.02 1023 = 4.09 1022
2a n(atoms) = = = 0.500 mol
b n(molecules) = = = 2.0 × 10–14 mol
c n(ions) = = × 3 = 6.23 × 10–4 mol
d n(atoms) = = × 24 = 0.24 mol
3 a Mr(C3H8) = 3 × 12.01 + 8 × 1.01 = 44.11b Mr(CH3NH2) = 12.01 + 5 × 1.01 + 14.01 = 31.07 c Mr(SO3) = 32.06 + 3 × 16.00 = 80.06d Mr(H2O2) = 2 × 1.01 + 2 × 16.00 = 34.02
4 a 239.19 gb 134.45 gc 98.08 gd 17.04 g
5a n(C) = = = 3.0 mol
b n(O3) = = = 0.20 mol
c n(SO3) = = = 7.7 10–4 mol
d n(V2O5) = = = 16.6 mol
6a n(BaSO4) = = = 1.07 mol
n(ions) = 2n(BaSO4) = 2.14 mol
b n(Fe2O3) = = = 3.13 103 mol
n(ions) = 5n(Fe2O3) = 1.57 × 104 mol
c n(KMnO4) = = = 5.86 10–3 mol
Page 1
© Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2008. This page from the Chemistry: For use with the IB Diploma Programme SL Teacher’s Resource may be reproduced for classroom use.
Worksheet 4.1: SolutionsMole calculations
No. Answer
n(ions) = 2n(KMnO4) = 1.17 10–2 mol
d n(Al2O3) = = = 2.94 105 mol
n(ions) = 5n(Al2O3) = 1.47 × 106 mol
7 a m(K2Cr2O7) = n M = 7.028 10–4 294.2 = 0.2068 gb m(C8H18) = n M = 40.0 114.26 = 4.57 × 103 gc m(C14H18N2O5) = n M = 2.80 10–5 294.34 = 8.24 10–3 gd m(Ca3(AsO4)2) = n M = 3.079 398.08 = 1226 ge m(CH3COOH) = n M = 0.0464 60.06 = 2.79 g
8a n(H) = n(HNO2) = = = 0.315 mol
b n(H) = 2n(BeH2) = 2 × = 2 × = 6.35 103 mol
c n(H) = 14n(C6H13OH) = 14 × = 14 × = 13.7 mol
d n(H) = 3n(Al2(HPO4)3) = 3 × = 3 × = 8.57 10–3 mol
9a m(C2H5OH) = M = × 46.08 = 1.91 × 10–7 g
b m(C6H8O7) = M = × 192.14 = 4.79 × 10–20 g
c m(H2) = M = × 2.02 = 5.03 g
d m(KMnO4) = M = × 158.04 = 189 g
e m(C8H18) = M = × 114.26 = 1.90 × 10–22 g
10M(XO2) = =
5537
0 693
.
. = 79.90 g mol–1
M(X + (2 16.00)) = 79.90 g mol–1
M(X) = 47.90 g mol–1, and so element X is titanium.
11 a M(C5H5NO2) = 111.11 g mol–1
b m(C5H5NO2) = n M = 8.22 10–4 111.11 = 0.0913 g
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© Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2008. This page from the Chemistry: For use with the IB Diploma Programme SL Teacher’s Resource may be reproduced for classroom use.
