Chemistry MCQs for NEET — Practice Questions with Answers

(c) Molar mass = 70 x 2 =140;

(CO)x, (12+16).x =140

therefore, x=5

(a)  2PH₃(g) $\to$2P(s) + 3H₂(g)

       100             0            0         Before dissociation

         0               -            150     After dissociation

More explanation below: 

As the volume occupied by gases only depends on number of moles of gases, the answer is related to number of moles of gases produced. 

Since, no temperature and pressure is given, one can assume NTP. 

At NTP, the reaction produces Phosphorous(Solid) and Hydrogen(gas). Looking at the balanced equation above, 2 moles of PH₃ gas would give 3 moles of H₂ gas. 

Therefore, there is 50% increase in number of gaseous molecules after reaction. (2 to 3 moles)

As the volume depends on number of moles, the increased volume is also 50% more than original volume. 

Increase in volume = Original Volume * 50 % = 100 ml * 50% = 50ml

New Volume = Original Volume + Increase in Volume = (100 + 50) ml = 150 ml 

(b) Mass of solvent = mass of solution - mass of NaCl

= 1.0585 x 1000 - 58.5

= 1058.5 - 58.5 = 1000g =1kg

m = mole of NaCl / mass of solvent in kg = 1/1 = 1

(b)  Mole of Cu²⁺ = 0.1x1 = 0.1

Mole of S${\mathrm{O}}_4^{2-}$=0.1x1=0.1

Mole of Al³⁺ = 0.1x2=0.2

Mole of S${\mathrm{O}}_4^{2-}$ = 0.1x3 = 0.3

Total moles of  ions present in1 litre =0.7

Molarity of all ions = 0.7M

In this redox titration, KMnO4 oxidizes As2O3 to H3AsO4. The balanced equation is: 2KMnO4 + 5As2O3 + 3H2SO4 → 2MnSO4 + 5H3AsO4 + K2SO4. Given: Mass of As2O3 = 0.1097 g, Volume of KMnO4 = 26.10 ml. Using the mole ratio from the balanced equation, we can calculate the molarity of KMnO4 as 0.018 M.

Solution approach 

CaCO₃ → CaO + CO₂ (decomposition reaction for 50% pure part on heating)

The impure part doesn't react as per the question, it remains as it is. So, total residue after heating will be 

1) mass of the impure part + mass of CaO (left from pure part) 

 since vol of CO₂         

 _{suggest that moles are 1.12/22.4 = 0.05 moles}

_{so wt of ${\mathrm{CaCO}}_3$ = 0.05 .100 = 5 gram. since 50 % is impure , so 5 gram is impure} 

You will get mass of impure part = 5g  & mass of CaO = 2.8 

So total = 5 + 2.8 = 7.8g 

3BaCl₂ + 2Na₃PO₄ $\to$Ba₃(PO₄)₂ +6NaCl

0.5mole 0.2 mole

Na₃PO₄ is the limiting reagent

2 Mole Na₃PO₄ $\to$ 2 Mole Ba₃(PO₄)₂ +6NaCl

0.2 Mole  Na₃PO₄ $\to$0.1 Mole Ba₃(PO₄)₂

1 molecule of oxygen = 32 a.m.u=32 x 1.66x 10^-24g

1 atom of nitrogen =14 a.m.u=14 x 1.66x 10^-24g

1 x 10^-10 g molecular weight of oxygen =  avg no of molecules of oxygen is 32 gram.so wt of given no is 

 32 x 10^-14g

1 x 10^-10g atomic weight of copper = 63.5 x 10^-14g

So, order is II < I <III <IV

No. of atoms in 2.2 g CO₂(n₁) = $\frac{22}{44}\times 3\times N_A$

No. of atoms in 1.7g NH₃(n₂) =$\frac{1.7}{17}\times 4\times N_A$

$\frac{n_1}{n_2}=\frac{3}{8}$