# Language of Chemistry Exe-1C Numericals Answer Chemistry Class-9 ICSE Selina Publishers

Language of Chemistry Exe-1C Numericals Answer Chemistry Class-9 ICSE Selina Publishers Solutions Chapter-1. Step By Step ICSE Selina Concise Solutions of Chapter-1 Language of Chemistry with All Exercise including MCQs, Very Short Answer Type, Short Answer Type, Long Answer Type, Numerical and Structured/Application Questions Solved . Visit official Website CISCE for detail information about ICSE Board Class-9.

## Language of Chemistry Exe-1C Numericals Answer Chemistry Class-9 ICSE Concise Selina Publishers

 Board ICSE Publications Selina Publication Subject Chemistry Class 9th Chapter-1 Language of Chemistry Book Name Concise Topics Solution of Exercise – (1C) Numericals Answer Type Academic Session 2023-2024

### F. Exercise – (1C) Numericals Answer Type

Language of Chemistry Class-9 Chemistry Concise Solutions

Page 21

#### Question 1.

Calculate the molecular mass of the following :

(i) Na2SO4.10H2O

(ii) (NH4)2CO3

(iii) (NH2)2CO

(iv) Mg3N2

Given atomic mass of Na = 23, H = 1, O = 16, C = 12, N = 14, Mg = 24, S = 32

(i) Molecular mass of Na2SO4.10H2O
= 2 x 23 + 32 + 4 x 16 + 10(2 x 1 + 16)
= 46 + 32 + 64 + 180
= 322 amu

(ii) Molecular mass of (NH4)2CO3
= 2(14 + 4 x 1) + 12 + 3 x 16
= 36 + 12 + 48
= 96 amu

(iii) Molecular mass of (NH2)2CO
= 2(14 + 2 x 1) + 12 + 16
= 32 + 12 + 16
= 60 amu

(iv) Molecular mass of Mg3N2
= 3 x 24 + 2 x 14
= 72 + 28
= 100 amu

#### Question 2.

Calculate the relative molecular masses of:

(a)CHCL3

(b) (NH4)2Cr2O7

(c) CuSO4.5H2O

(d) (NH4)2SO4

(e) CH3COONa

(f) potassium chlorate

(g) Ammonium chloroplatinate =(NH4)2 SO4

[At. Mass: C = 12, H =1, O = 16, Cl = 35.5, N = 14, Cu = 63.5, S =32, Na = 23, K = 39, Pt = 195, Ca = 40, P =31 Mg = 24,]

(a) The relative molecular mass of CHCl3
= 12 + 1 + 3 x 35.5
= 12 + 1 + 106.5
119.5 amu

(b) The relative molecular mass of (NH4)2Cr2O7
= 2(14 + 4 x 1) + 2 x 52 + 7 x 16
= 36 + 104 + 112
252 amu

(c) The relative molecular mass of CuSO4.5H2O
= 63.5 + 32 + 4 x 16 + 5(2 x 1 + 16)
= 63.5 + 32 + 64 + 90
249.5 amu

(d) The relative molecular mass of (NH4)2SO4
= 2(14 + 4 x 1) + 32 + 4 x 16
= 36 + 32 + 64
132 amu

(e) The relative molecular mass of CH3COONa
= 12 + 3 + 12 + 16 + 16 + 23
82 amu

(f) The relative molecular mass of KClO3
= 39 + 35.5 + 3 x 16
= 39 + 35.5 + 48
122.5 amu

(g) The relative molecular mass of (NH4)2PtCl6
= 2(14 + 4 x 1) + 195 + 6 x 35.5
= 36 + 195 + 213
444 amu

#### Question 3.

Find the percentage mass of water in Epsom salt, MgSO4.7H2O.

Relative molecular mass of MgSO4.7H2O
= 24 + 32 + 4 x 16 + 7(2 x 1 + 16)
= 24 + 32 + 64 + 126
= 246 amu

246 g of Epsom salt contains 126 g of water of crystallisation

∴ 100 g of Epsom salt contain (126 x 100)/246

= 12600/246 = 51.22 g of water of crystallisation.

∴ % mass of water in Epsom salt, MgSO4.7H2O = 51.22%

#### Question 4.

Calculate the percentage of phosphorus in:

(a) Calcium hydrogen phosphate, Ca(H2PO4)2

(b) Calcium phosphate, Ca3(PO4)2

(a) Relative molecular mass of Ca(H2PO4)2
= 40 + 2(2 x 1 + 31 + 4 x 16)
= 40 + 2(2 + 31 + 64)
= 40 + 194
= 234 amu

Wt. of P in Ca(H2PO4)2 = 2 x 31 = 62 g

% of P = Wt. of P/Total Wt. of Ca(H2PO4)2×100

= 62/234 x 100 = 26.5%

∴ Phosphorus in Calcium hydrogen phosphate is 26.5%

(b) Relative molecular mass of Calcium phosphate, Ca3(PO4)2
= 3 x 40 + 2(31 + 4 x 16)
= 120 + 2(31 + 64)
= 120 + 190
= 310 amu

Wt. of P in Ca3(PO4)2 = 2 x 31 = 62 g

% of P = Wt. of P/Total Wt. of Ca3(PO4)2×100

= 62/310 x 100 = 20%

∴ Phosphorus in Calcium phosphate is 20%

#### Question 5.

Calculate the percentage composition of each element in Potassium chlorate, KClO3.

Relative molecular mass of KClO3
= 39 + 35.5 + 3 x 16
= 39 + 35.5 + 48
= 122.5 amu

122.5 g of KClO3 contains 39 g of Potassium

∴ 100 g of KClO3 contains 39×100/122.5 g of Potassium

= 39000/1225 = 31.83 g of Potassium

122.5 g of KClO3 contains 35.5 g of Chlorine

122.5 g of KClO3 contains 35.5 g of Chlorine

∴ 100 g of KClO3 contains 35.5×100/122.5 g of Chlorine

= 35500/1225 = 28.98 g of Chlorine

122.5 g of KClO3 contains 48 g of Oxygen

∴ 100 g of KClO3 contains 48×100/122.5 g of Oxygen

= 48000/1225 = 39.18 g of Oxygen

∴ In KClO3 : K = 31.83%, Cl = 28.98% and O = 39.18%

#### Question 6.

Urea is a very important nitrogenous fertilizer. Its formula is CON2H4. Calculate the percentage of carbon in urea.

(C= 12, O = 16, N = 14 and H = 1)

Relative molecular mass of CON2H4
= 12 + 16 + 2 x 14 + 4 x 1
= 12 + 16 + 28 + 4
= 60 amu

Wt. of C in CON2H4 = 12 g

% of C = Wt. of C/Total Wt. of CON2H4)×100

= 12/60 x 100 = 20%

∴ Carbon in Urea is 20%

—  : End of Language of Chemistry Exe-1C Numericals Answer Class-9 ICSE Chemistry Solutions :–

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