A chloride of a metal (M) has 65.5% of chlorine. 100 ml of vapour of the chloride of metal at STP weighs 0.72 g. The molecular formula of this metal chloride is

1. MCl₃ 

2. MCl₄ 

3. M₂Cl₃ 

4. MCl₅ 

The molality of 0.01 M Na₂CO₃ solution is:
(The density of the solution is 1.1 g/ml)

1. $9\times {10}^{-3}$

2. $1.8\times {10}^{-3}$

3. $4.5\times {10}^{-3}$

4. $1.1\times {10}^{-3}$

The weight of 1x10²³ molecules of CuSO₄.5H₂O is: 

1. 34.42 g

2. 41.42 g

3. 54.44 g

4. 68.94 g

lf N_A is Avogadro's number, then the number of valence electrons in 4.2 g of nitride ions (N${}^{3-}$) will be:

ln the final answer of the expression \(\large{\frac{(29.2-20.2)\times(1.79\times10^5)}{1.37}}\)
the number of significant figures after solving the expression is:

When 7.3 g of magnesium bicarbonate (Mg(HCO₃)₂) is decomposed by heating, how much carbon dioxide gas (in mL) is released at STP?

1. 1000 mL

2. 1120 mL

3. 2230 mL

4. 3240 mL

The amount of Zinc (atomic weight = 65) necessary to produce 224 mL of H₂ (at STP) by the reaction with dilute sulphuric acid will be: 

On complete decomposition, the volume of CO₂ released at STP on heating 9.85 g of BaCO₃ (atomic mass, Ba = 137) will be:

1. 1.12 L

2. 4.84 L

3. 2.12 L

4. 2.06 L

Specific volume of cylindrical virus particle is $6.02\times {10}^{-2} \mathrm{cc}/\mathrm{g}.$ Whose radius and length are 7 Å and 10 Å respectively. If ${\mathrm{N}}_{\mathrm{A}}=6.02\times {10}^{23},$ the molecular weight of virus is -

1. $15.4 \mathrm{kg}/\mathrm{mol}$

2. $1.54\times {10}^4 \mathrm{kg}/\mathrm{mol}$

If $3.02\times {10}^{19}$ molecules are removed from 98 mg of H₂SO₄, then the number of moles of H₂SO₄ left are -

1. $0.1\times {10}^{-3}$

2. $9.5\times {10}^{-4}$

3. $1.2\times {10}^{-4}$

4. $0.5\times {10}^{-3}$