(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.

Question

(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.

KnowledgeBoat · Accepted 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 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 = 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: = 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 = Volume occupied by 2 g of O₂ gas at 27°C and 740 mm pressure: S.T.P. | Given Values --------|-------------- P₁ = 760 mm of Hg | P₂ = 740 mm of Hg V₁ = 1.4 lit | V₂ = x lit T₁ = 273 K | T₂ = 27 + 273 K Using the gas equation, Substituting the values we get, 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 Values | S.T.P. --------|-------------- P₁ = 750 mm of Hg | P₂ = 760 mm of Hg V₁ = 44 lit | V₂ = x lit T₁ = 25 + 273 K = 298 K| T₂ = 273 K Using the gas equation, Substituting the values we get, 22.4 litre of CO₂ at S.T.P. has mass = 44 g 39.78 litre of CO₂ at S.T.P. has mass x 39.78 = 78.14 g (e) (i) 143.5 g AgCl is formed by 58.5 g NaCl 1.435 g of AgCl will be formed by x 1.435 = 0.582 g Percentage of NaCl = x 100 = 58.5% Hence, percentage of NaCl is 58.5%

Chemistry

Mole Concept

Related Questions

From the equation:
C + 2H₂SO₄ ⟶ CO₂ + 2H₂O + 2SO₂
Calculate:
(i) The mass of carbon oxidized by 49 g of sulphuric acid.
(ii) The volume of sulphur dioxide measured at STP, liberated at the same time.

Chemistry

Mole Concept

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From the equation:C + 2H2SO4 ⟶ CO2 + 2H2O + 2SO2Calculate:(i) The mass of carbon oxidized by 49 g of sulphuric acid.(ii) The volume of sulphur dioxide measured at STP, liberated at the same time.

From the equation:
C + 2H₂SO₄ ⟶ CO₂ + 2H₂O + 2SO₂
Calculate:
(i) The mass of carbon oxidized by 49 g of sulphuric acid.
(ii) The volume of sulphur dioxide measured at STP, liberated at the same time.