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A mixture contains NaCI and unknown chloride MCI.
(i) 1 g of this is dissolved in water. Excess of acidified AgNO3 solution is added to it. 2.567 g of white ppt. is formed.
(ii) 12 g of original mixture is heated to 300oC. Some vapour come out which are absorbed in acidified AgNO3 solution, 1.341 g of white precipitate was obtained. Find the molecular weight of unknown chloride.

Hrishant Goswami , 11 Years ago
Grade 10
anser 1 Answers
Jitender Pal
Weight of AgCI formed = 2.567 g
Amount of AgCI formed due to MCI = 1.341 g
(∵ NaCI does not decompose on heating to 300oC)
∴ Weight of AgCI formed due to NaCI
= 2.567 – 1.341 = 1.226g
NaCI ≡ AgCI ≡ MCI
58.5 143.5
{NaCI + AgNO3 → AgCI + NaNO3 MCI + AgNO3 → AgCI + MNO3}
∵ 143.5g of AgCI is obtained from NaCI = 58.5g
∴ 1.226 g of AgCI is obtained form NaCI
= 58/143.5 * 1.226 = 0.4997 g
∴Wt of MCI 1 g of mixture = 1.000 – 0.4997 = 0.5003g
∵ 1.341 g of AgCI is obtained from MCI = 0.5003g
∴ 143.5g of AgCI is obtained from MCI
= 0.5003/1.341 * 143.5 * 53.53 g
∴ Molecular weight of MCI = 53.53
Last Activity: 11 Years ago
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