Chemistry
1 gram mixture of NaCl and NaNO3 is dissolved in water and treated with an excess of AgNO3 solution. 1.43 g of precipitate is formed. Calculate percentage of sodium chloride in the mixture. [5]
[Ag = 108, Na = 23, N = 14, O = 16, Cl = 35.5].
Stoichiometry
5 Likes
Answer
Molecular mass of AgCl = 108 + 35.5 = 143.5
Molecular mass of NaCl = 23 + 35.5 = 58.5
143.5 g AgCl is formed by 58.5 g NaCl
∴ 1.43 g of AgCl will be formed by x 1.43 = 0.58 g
Percentage of NaCl = x 100 = 58.29%
Hence, percentage of NaCl is 58.29%
Answered By
2 Likes
Related Questions
Give equations for the following :
(i) Preparation of acetylene
(ii) Preparation of ethyl alcohol
(iii) Addition of chlorine to ethylene
(iv) Preparation of ethane
(v) Sulphurous acid reacts with bromine.
Complete and balance the following:
(i) Potassium nitrate + sulphuric acid ⟶
(ii) Ferrous chloride + Ammonium hydroxide ⟶
(iii) Red lead + conc. HCl ⟶
(iv) Sodium thiosulphate + dilute Hydrochloric acid ⟶
(v) Copper with dil nitric acid ⟶
[Dilute sulphuric acid, copper, iron, sodium, zinc, copper carbonate and sodium carbonate]
Choosing only from the list of substances mentioned above, write an equation for the reaction which you would use in the lab to obtain :
(i) Zinc carbonate
(ii) Copper sulphate
(iii) Sodium sulphate
(iv) Iron sulphate
A volatile chloride of metal M contains 34.5% of the metal. If the density of metal chloride relative to hydrogen is 162.5. Calculate the molecular formula of metal chloride.
[M = 56, Cl = 35.5].