Experimentally it was found that a metal oxide has formula M_0.98O. Metal M, is present as M²⁺ and M³⁺ in its oxide. The fraction of the metal which exists as M³⁺ would be:

• 7.01%

• 4.08%

• 6.05%

• 6.05%

The most abundant element by mass in the body of a healthy human adult are :
Oxygen (61.4%) ; Carbon (22.9%), Hydrogen(10.0%) ; and Nitrogen (2.6%). The weight which a 75 kg person would gain if all 1H atoms are replaced by 2H atoms is

• 15 kg

• 37.5 Kg

• 7.5 Kg

• 7.5 Kg

How many moles of magnesium phosphate, Mg₃(PO₄)² will contain 0.25 mole of oxygen atoms?

• 0.02

• 3.125 × 10^–2

• 1.25 × 10^–2

• 1.25 × 10^–2

264.

If we consider that 1/6 in place of 1/12 mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will 

• Decrease twice

• Increase two fold

• Remain unchanged

• Remain unchanged

The ratio of mass percent of C and H of an organic compound (C_XH_YO_Z) is 6: 1. If one molecule of the above compound (C_XH_YO_Z) contains half as much oxygen as required to burn one molecule of compound C_XH_Y completely to CO₂ and H₂O. The empirical formula of compound C_XH_YO_Z is

• C₂H₄O₃

• C₃H₆O₃

• C₂H₄O

• C₃H₄O₂

0.126 g of an acid is needed to completely neutralise 20 mL 0.1 (N) NaOH solution. The equivalent weight of the acid is

• 53

• 40

• 45

• 63

In a flask, the weight ratio of CH₄ (g) and SO₂ (g) at 298K and 1 bar is 1 : 2. The ratio of the number of molecules of SO₂ (g) and CH₄ (g) is

• 1 : 4

• 4 : 1

• 1 : 2

• 2 : 1

You are supplied with 500 mL each of 2N HCl and SN HCL What is the maximum volume of 3M HCl that you can prepare using only these two solutions?

• 250 mL

• 500 mL

• 750 mL

• 1000 mL

What will be the normality of the salt solution obtained by neutralising x mL y (N) HCl with y mL x(N) NaOH, and finally adding (x + y) mL distilled water

• $\frac{2(\mathrm{x} + \mathrm{y})}{\mathrm{xy}} \mathrm{N}$

• $\frac{\mathrm{xy}}{2(\mathrm{x} + \mathrm{y})}\mathrm{N}$

• $\left(\frac{2\mathrm{xy}}{\mathrm{x} + \mathrm{y}}\right) \mathrm{N}$

• $\left(\frac{\mathrm{x} + \mathrm{y}}{\mathrm{xy}}\right)\mathrm{N}$

At a particular temperature, the ratio of equivalent conductance to specific conductance of a 0.01 N NaCl solution is

• 10⁵ cm³

• 10³ cm³

• 10 cm³

• 10⁵ cm²