Find the percentage of boron in Na 2 B 4 O 7 · 10 H 2 O . ( H = 1 , B = 11 , O = 16 , Na = 23 ).

Percentage of an element in a compound = Total weight of the element in one molecule Gram molecular weight of the compound × 100 Relative molecular mass of Na 2 B 4 O 7 · 10 H 2 O = 2 × 23 + 4 × 11 + 7 × 16 + 10 × 18 = 382 382 g of Na 2 B 4 O 7 · 10 H 2 O contains 44 g of boron, Therefore, the percentage of boron = 44 382 × 100 = 11 . 5 %

Solid ammonium dichromate decomposes as NH 4 2 Cr 2 O 7 → Cr 2 O 3 + N 2 + 4 H 2 O , if 63 g of ammonium dichromate decomposes, calculate the volume of N 2 evolved at STP .

Consider the given reaction, NH 4 2 Cr 2 O 7 → Cr 2 O 3 + N 2 + 4 H 2 O . Molecular mass of NH 4 2 Cr 2 O 7 = 2 × 14 + 4 + 2 × 52 + 7 × 16 = 252 We know that, the number of moles = Mass of the substance Molecular mass = 63 252 = 0 . 25 mol . According to the above balanced equation, the decomposition of one mole of ammonium dichromate gives one mole of nitrogen gas. Therefore, the number of moles of nitrogen gas formed from the decomposition of 0 . 25 mol of ammonium dichromate = 0 . 25 mol . We know that, one mole of any gas occupies 22 . 4 L at STP , therefore, the volume of nitrogen gas evolved = 22 . 4 × 0 . 25 = 5 . 6 L .

Solid ammonium dichromate decomposes as NH 4 2 Cr 2 O 7 → Cr 2 O 3 + N 2 + 4 H 2 O , if 63 g of ammonium dichromate decomposes, calculate the loss of mass.

Consider the given reaction, NH 4 2 Cr 2 O 7 → Cr 2 O 3 + N 2 + 4 H 2 O Molecular mass of ammonium dichromate NH 4 2 Cr 2 O 7 = 2 × 14 + 4 + 2 × 52 + 7 × 16 = 252 . Molecular mass of chromium oxide = 2 × 52 + 3 × 16 = 152 252 g of solid ammonium dichromate decomposes to give 152 g of solid chromium oxide, so the loss in mass in terms of solid formed = 252 - 152 = 100 g . If 63 g of ammonium dichromate is decomposed, the loss in mass = 63 × 100 252 = 25 g .

If 150 cc of gas A contains X molecules, how many molecules of gas B will be present in 75 cc of B ? The gases A and B are under the same conditions of temperature and pressure. Name the law on which the problem is based.

According to Avogadro’s law, equal volumes of all gases under similar conditions of temperature and pressure contain an equal number of molecules. So 150 cc of B will also contain X molecules. Hence, the number of molecules present in 75 cc of B will contain X 2 molecules.

Calculate the number of moles in 12 g of oxygen gas. [ O = 16 ]

Molecular mass = 16 × 2 = 32 g 32 g of oxygen gas → 1 mole 1 g of oxygen gas → 1 32 mole Given that 12 g of oxygen gas Number of moles = 12 32 = 0 . 375 mole