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To calculate the amount of analytical grade potassium dichromate required to prepare a 0.1 N solution, we need to use the following formula: Amount (in grams) = (Normality × Equivalent weight × Volume) / 1000 The equivalent weight of potassium dichromate (K₂Cr₂O_7) can be calculated by considering its molecular weight and the number of equivalents of potassium dichromate present in the formula. The formula weight of K₂Cr₂O_7 is: (2 × atomic mass of K) + (2 × atomic mass of Cr) + (7 × atomic mass of O) = (2 × 39) + (2 × 52) + (7 × 16) = 294 + 104 + 112 = 510 The equivalent weight is then calculated by dividing the formula weight by the number of equivalents. Since potassium dichromate donates two equivalents of Cr₂O_7^2- per mole, the equivalent weight is: Equivalent weight = Formula weight / Number of equivalents = 510 / 2 = 255 g/equivalent Now we can substitute the values into the formula: Amount (in grams) = (0.1 N × 255 g/equivalent × 1000 mL) / 1000 Simplifying further: Amount (in grams) = 0.1 × 255 = 25.5 grams Therefore, the correct answer is option 3: 4.9 grams.
2048 × 1824 ÷ 76 = (? - 212) × 64
22 + 60 × 3 ÷ 12 = ?
17% of 250 + ? = 108
(5/8 of 480 - 30% of 420)² ÷ (√81 + 25% of 320) = ?
(1296 ÷ 54 × 24 + 24) = ? × 150
(18 2 – 17 2) x (1/5) + ? = 148
Find the value of the given expression.
[76 – {90 ÷ 5 × (24 – 36 ÷ 3) ÷ 3}]
∛857375 + ∛91125 = ? + √6889
20 × 224÷ 16 – 50 = ? + 100
...2850 ÷ 2.5 - ? × 42 = 300