A gas is found to have a formula [CO]_x. If its vapour density is 70, the value of x is:

Solution

As we know that,Molecular mass of [CO]_x= 2 × V.D.

⇨(12 + 16)x = 2 ×70 ⇨ 28x = 140 ⇨x =5

If 1¹/₂ moles of oxygen combine with Al to form Al₂O₃the weight of Al used in the reaction is (Al = 27)

27 g
54 g
49.5 g
31 g

Solution

In the reaction
4NH₃(g) + 5O₂(g) ⟶ 4NO(g) + 6H₂O(l), when 1 mole of ammonia and 1 mole of O₂ are made to react to completion

1.0 mole of H₂O is produced
1.0 mole of NO will be produced
all the ammonia will be consumed
all the oxygen will be consumed

Solution

In order to prepare one litre normal solution of KMnO₄, how many grams of KMnO₄ are required if the solution is to be used in acid medium for oxidation?

158 g
62.0 g
31.6 g
790 g

Solution

Eq. wt of KMnO₄ in acid medium is 31.6 g. Hence this much amount must be dissolved in 1 litre to prepare normal solution.

An organic compound whose empirical and molecular formula are same, contains 20% carbon, 6.7% hydrogen,46.7%nitrogen and the rest oxygen. On heating it yields ammonia,leaving a solid residue. The solid residue gives a violet colour with dilute solution of alkaline copper sulphate. The organic compound is

NH₂COONH₄
HCOONH₄
NH₂NHCHO
NH₂CONH₂

Solution

A compound contains 54.55 % carbon, 9.09% hydrogen ,36.36% oxygen. The empirical formula of this compound is :

C₃H₅O
C₄H₈O₂
C₂H₄O₂
C₂H₄O

Solution

The mass of BaCO₃ produced when excess CO₂ is bubbled through a solution of 0.205 mol Ba(OH)₂ is:

81 g
40.5 g
20.25 g
162 g

Solution

The empirical formula of a compound is CH₂. One mole of this compound has a mass of 42 grams. Its molecular formula is :

C₃H₆
C₃H₈
CH₂
C₂H₂

Solution

Empirical formula of compound = CH₂ Molecular mass of the compound = 42
∴ n = 42/14 = 3
∴ Hence molecular formula = C₃H₆

30 g of magnesium and 30 g of oxygen are reacted, then the residual mixture contains

50 g of Magnesium oxide and 10 g of oxygen
40 g of Magnesium oxide and 20 g of oxygen
45 g of Magnesium oxide and 15 g of oxygen
60 g of Magnesium oxide only

Solution

The number of atoms in 4.25 g of NH₃ is approximately

6 × 10₂₃
2 × 10₂₃
4 × 10₂₃
1 × 10₂₃

Solution

Number of atoms = \(\frac{4.25 \times 6.023 \times 10^{4} \times 4}{17}\) = 6 × 10²³
(One molecule of NH₃ contains 4 atoms 1 N and 3H)

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