The maximum number of molecules are present in

15L of H₂ gas at STP
5L of N₂ gas at STP
0.5 g of H₂ gas
10g of O₂ gas

Solution

The equivalent weight of a solid element is found to be 9. If the specific heat of this element is 1.05 Jg^–1 K^–1, then its atomic weight is :

Solution

The molar solution of H₂SO₄ is equal to :

N/2 solution
N solution
2N solution
3N solution

Solution

What is the molarity of 0.2N Na₂CO₃ solution?

0.1 M
0 M
0.4 M
0.2 M

Solution

The equivalent weight of MnSO₄ is half of its molecular weight when it is converted to :

Mn₂O₃
MnO₂
MnO₄^-
MnO₄^2-

Solution

For equivalent weight of MnSO₄ to be half of its molecular weight, change in oxidation state must be equal to 2. It is possible only when oxidation state of Mn in product is + 4. Since oxidation state of Mn in MnSO₄2 is correct answer.
In MnO₂, O.S. of Mn = +4
∴Change in O.S. of Mn = +4 –(+2) = +2

Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is

2.28 mol kg^–1
0.44 mol kg^–1
1.14 mol kg^–1
3.28 mol kg^–1

Solution

To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H₃PO₃), the value of 0.1 M aqueous KOH solution required is

40 mL
20 mL
10 mL
60 mL

Solution

N₁V₁ = N₂V₂ (Note:H₃PO₃ is dibasic ∴ M = 2N)

20 × 0.2 = 0.1 × V (Thus. 0.1 M = 0.2 N)
∴ V = 40 ml

9 mol of CO₂
16 mol of CO₂
17 mol of CO₂
18 mol of CO₂

Solution

What volume of hydrogen will be liberated at NTP by the reaction of Zn on 50 ml dilute H₂SO₄ of specific gravity 1.3 and having purity 40%?

3.5 litre
8.25 litre
6.74 litre
5.94 litre

Solution

Two solutions of a substance (non electrolyte) are mixed in the following manner.480 ml of 1.5 M first solution+ 520 ml of 1.2 M second solution. What is the molarity of the final mixture?

2.70 M
1.344 M
1.50 M
1.20 M

Solution

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