NEET Practice Questions (MCQs) with Answers & Solutions

(B).    Case I : Mean molar mass =$\frac{16\mathrm{x}+28\mathrm{y}}{\mathrm{x}+\mathrm{y}}=20 \left(\mathrm{given}\right)$

$16\mathrm{x}+28\mathrm{y} = 20\mathrm{x}+20\mathrm{y} \mathbf{:} \mathrm{y}=0.5\mathrm{x}.$

         Case II : Mean molar mass

$=\frac{16\mathrm{x}+28\mathrm{y}}{\mathrm{x}+\mathrm{y}}=\frac{8\mathrm{x}+28\mathrm{x}}{\mathrm{x}+0.5\mathrm{x}}=\frac{36}{1.5}=24$

2.

Number of molecules = mole × avogadro number                 
= 0.1 × 6.022 × 10²³                  

= 6.022 × 10²²

1 molecule of PO43− contains 15 + 4 × 8 + 3 = 50 electron

So,

6.022 × 10²² contains 

= 50 × 6.022 × 10²²        

= 301.1 × 10²² electrons

$$\begin{gathered} \mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)=\frac{54}{18} \\ \left(\mathrm{d}=1.0 \mathrm{g}/\mathrm{mL} \mathrm{for} {\mathrm{H}}_2\mathrm{O}\right) \\ \mathrm{H} \mathrm{has} \mathrm{no} \mathrm{neutron} \\ \therefore \mathrm{no}. \mathrm{of} \mathrm{neutrons} \mathrm{in} {\mathrm{H}}_2\mathrm{O}=3\times 8\times {\mathrm{N}}_{\mathrm{A}} \\ = 24 {\mathrm{N}}_{\mathrm{A}} \end{gathered}$$

Molar gas volume at STP = 22.4 litres

We need to convert this volume to the volume the gas would occupy at STP

p1 = 8.21 atm                                  p2 = 1 atm

v1 = 3 litres                                      v2 = ?

t1 = 27 + 273 = 300 k                       t2 = 273K

p1v1/t1 = p2v2/t2

(8.21x3)/300 = (1xv2)/273

From which

v2 = 22.4133 litres

Number of moles of O3 = 22.4133/22.4 = 1 mole

Each O3 molecule has 3 atoms of O

Number of moles of oxygen atoms =  3 x 1

                                         = 3

3.

If two different elements combine separately with the same weight of a third element, the ratio of the masses in which they do so are either the same or a simple multiple of the mass ratio in which they combine

for fixed wt of A ( 12 gram) the ratio of wt of B is 1;2 , so law of multiple proportion.

Calculate the mass and moles of H2O driven off: 
6.85 g - 3.13 g = 3.72 g H2O / 18.0 g/mol = 0.207 moles H2O
2- Calculate moles Sr(OH)2 remaining: 
3.13 g Sr(OH)2 / 121.6 g/mol = 0.0257 moles Sr(OH)2
3- Calculate the ratio of moles H2O / moles Sr(OH)2: 
0.207 / 0.0257 = 8.0
so, here the value of x will turns out to be 8 .

$$\begin{gathered} \mathrm{Wt}. \mathrm{of} {\mathrm{S}}_8 \mathrm{in} \mathrm{sample} = 160 \mathrm{g}; \\ \mathrm{Moles} \mathrm{of} {\mathrm{S}}_8=\frac{160}{32\times 8}=0.625 \\ \mathrm{No}. \mathrm{of} \mathrm{moles} \mathrm{of} {\mathrm{O}}_2 \mathrm{required} = 0.625\times 8 \\ \mathrm{Vol}. \mathrm{of} {\mathrm{O}}_2 \mathrm{required} \mathrm{at} \mathrm{STP} = 22.4\times 5 \\ \therefore \mathrm{Vol}. \mathrm{of} \mathrm{air} \mathrm{required} \mathrm{at} \mathrm{STP} = 22.4\times 5\times \frac{100}{21}= 533.33 \mathrm{L} \end{gathered}$$

$$\begin{gathered} \mathrm{Produced} \mathrm{mass} \mathrm{of} {\mathrm{H}}_3{\mathrm{PO}}_4=\left(\frac{62}{4\times 31}\right)\times 0.85\times 0.9\times 4\times 98 \\ = 149.94 \mathrm{g} \end{gathered}$$

$$\begin{gathered} \mathrm{Total} \mathrm{moles} \mathrm{of} {\mathrm{H}}_2{\mathrm{SO}}_4=0.1 \mathrm{mole} \\ \mathrm{Total} \mathrm{volume}=\frac{150+400}{1.25}=\frac{550}{1.25}=440 \\ \therefore \mathrm{M}=\frac{0.1}{440}\times 1000=\frac{1}{4.4}=0.227 \mathrm{M} \end{gathered}$$