Bright Tutorials Bright Tutorials
ICSE Class 10 Chemistry Question 20 of 72

Mole Concept & Stoichiometry Miscellaneous Exercises — Question 20

Back to all questions
20
Question

Question 20

(a) A hydrate of calcium sulphate CaSO4.xH2O contains 21% water of crystallisation. Find the value of x.

(b) What volume of hydrogen and oxygen measured at S.T.P. will be required to prepare 1.8 g of water.

(c) How much volume will be occupied by 2g of dry oxygen at 27°C and 740 mm pressure?

(d) What would be the mass of CO2 occupying a volume of 44 litres at 25°C and 750 mm pressure?

(e) 1 g of a mixture of sodium chloride and sodium nitrate is dissolved in water. On adding silver nitrate solution, 1.435 g of AgCl is precipitated.

AgNO3 (aq) + NaCl (aq) ⟶ AgCl (s) + NaNO3

Calculate the percentage of NaCl in the mixture.

Answer

Relative molecular mass of CuSO4.xH2O

= 40 + 32 + (4×16) + [x(2+16)]

= 40 + 32 + 64 + 18x

= 136 + 18x

∴ 21% water of crystallization = 18 x

⇒18x136+18x=211001800x=21(136+18x)1800x=2856+378x1800x−378x=28561422x=2856x=28561422x=2\Rightarrow \dfrac{18x}{136 + 18x } = \dfrac{21}{100} \\[1em] 1800x = 21(136 + 18x) \\[1em] 1800x = 2856 + 378x \\[1em] 1800x - 378x = 2856 \\[1em] 1422x = 2856 \\[1em] x = \dfrac{2856}{1422} \\[1em] x = 2

Hence, water of crystallization = 2

(b) Molar mass of H2O = 2(1) + 16 = 18 g

For 18 g water, vol. of hydrogen needed = 22.4 litre

∴ For 1.8 g, vol. of H2 needed = 22.418\dfrac{22.4}{18} x 1.8 = 2.24 L

According to Gay Lussac's Law, 2 vol of hydrogen requires 1 vol of oxygen

When 2 vol of hydrogen in 1.8 g H2O is 2.24 L, then one vol. of oxygen will be:

2.242\dfrac{2.24}{2} = 1.12 L

(c) Gram molecular mass of O2 = 2 x 16 = 32 g

1 mole of O2 weighs 32 g and occupies 22.4 lit. vol.

∴ 2 g of O2 occupies = 22.432×2=1.4 lit.\dfrac{22.4}{32} \times 2 = 1.4 \text{ lit.}

Volume occupied by 2 g of O2 gas at 27°C and 740 mm pressure:

S.T.P.Given Values
P1 = 760 mm of HgP2 = 740 mm of Hg
V1 = 1.4 litV2 = x lit
T1 = 273 KT2 = 27 + 273 K

Using the gas equation,

P1V1T1=P2V2T2\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}

Substituting the values we get,

760×1.4273=740×x300x=760×1.4×300740×273x=3,19,2002,02,020x=1.58 lit\dfrac{760 \times 1.4}{273} = \dfrac{740 \times x}{300} \\[0.5em] x = \dfrac{760 \times 1.4 \times 300}{740 \times 273 } \\[0.5em] x = \dfrac{3,19,200}{2,02,020} \\ \\[0.5em] x = 1.58 \text{ lit}

Hence, the volume occupied by 2 g of O2 gas at 27°C and 740 mm pressure is 1.58 lit.

(d) Gram molecular mass of CO2 = 12 + 2(16) = 12 + 32 = 44 g

Given ValuesS.T.P.
P1 = 750 mm of HgP2 = 760 mm of Hg
V1 = 44 litV2 = x lit
T1 = 25 + 273 K = 298 KT2 = 273 K

Using the gas equation,

P1V1T1=P2V2T2\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}

Substituting the values we get,

750×44298=760×x273x=750×44×273760×298x=9,009,000226,480x=39.78 lit\dfrac{750 \times 44}{298} = \dfrac{760 \times x}{273} \\[0.5em] x = \dfrac{750 \times 44 \times 273}{760 \times 298 } \\[0.5em] x = \dfrac{9,009,000}{226,480} \\ \\[0.5em] x = 39.78 \text{ lit}

22.4 litre of CO2 at S.T.P. has mass = 44 g

39.78 litre of CO2 at S.T.P. has mass 4422.4\dfrac{44}{22.4} x 39.78

= 78.14 g

(e)

AgNO3+NaCl⟶AgCl+NaNO323+35.5108+35.5=58.5g=143.5g\begin{matrix} \text{AgNO}_3 & + &\text{NaCl} & \longrightarrow & \text{AgCl} & + & \text{NaNO}_3 \\ && 23 + 35.5 && 108 + 35.5 \\ && = 58.5 \text{g}&& = 143.5\text{g} \\ \end{matrix}

(i) 143.5 g AgCl is formed by 58.5 g NaCl

∴ 1.435 g of AgCl will be formed by 58.5143.5\dfrac{58.5}{143.5} x 1.435 = 0.582 g

Percentage of NaCl = 0.5821\dfrac{0.582}{1} x 100 = 58.5%

Hence, percentage of NaCl is 58.5%