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A hydrated salt, Na 2 SO 4. nH 2 O undergoes 56% loss in weight on heating and become anhydrous. The value of n(approximately) will be : (PLEASE ANSWER IN A EASY TO UNDERSTAND MANNER)
8 Ιουλ 2024 · Here, \( Na2SO4 \) is sodium sulfate and \( nH2O \) represents the water of hydration. Step 2: Calculate the molar mass of the components - The molar mass of \( Na2SO4 \): - Sodium (Na): \( 22.99 \, g/mol \times 2 = 45.98 \, g/mol \) - Sulfur (S): \( 32.07 \, g/mol \) - Oxygen (O): \( 16.00 \, g/mol \times 4 = 64.00 \, g/mol \) Total for ...
1 Ιουν 2019 · The hydrated salt, `Na_(2)SO_(4)*nH_(2)O` undergoes `55.9%` loss in weight on heating and becomes anhydrous. The value of `n` will be: A. `5` B. `3` C. `7` D. `10`
7 Ιουν 2024 · The water of crystallisation is separated from the main formula by a dot when writing the chemical formula of hydrated compounds. E.g. hydrated copper (II) sulfate is CuSO 4∙ 5H 2 O. A compound which doesn’t contain water of crystallisation is called an anhydrous compound. E.g. anhydrous copper (II) sulfate is CuSO 4.
By solving this we will get $ n = 10 $ . The hydrated salt was $ N{a_2}S{O_4}.10{H_2}O $ . So, option D is correct. Note Sodium sulphate is used as a thermal storage.
2 Νοε 2023 · Salts that contain water within their structure are called hydrated salts. Anhydrous salts are those that contain no water in their structure. A common example is copper (II) sulfate which crystallises forming the salt hydrated copper (II) sulfate, which is blue.
detailed solution. Correct option is D. Loss in weight is due to nH 2 O. ∴ 142 + 18 n g of Na 2 SO 4. nH 2 O = 18 ng loss in weight of H 2 O. 100 g of Na 2 SO 4. nH 2 O = 18 n × 100 142 + 18 n. ∴ 18 n × 100 142 + 18 n = 55. 9. Solve for n ⇒ n = 9. 99 ≈ 10.
