Chemistry MCQs for NEET — Practice Questions with Answers

(a) 1.00 molal aqueous solution = 1.0 mole in 1000 g water n_solute = 1; W_solute = 1000 g.

n_solvent = 1000/18 =55.56

X_solute = 1/(1+55.56) = 0.0177

Molarity = number of moles of solute/volume of solution(in mL) x 1000

            = (25.3x1000)/(106x250) = 0.9547 $\approx$0.955 M

Na₂CO₃ in aqueous solution remains dissociated as

        Na₂CO₃$\to$2Na⁺ + ${\mathrm{CO}}_3^{2-}$

           x                  2x               x

Since, the molarity of Na₂CO₃ is 0.955 M, the molarity of ${\mathrm{CO}}_3^{2-}$ is also 0.955 M and that of Na⁺ is

2 x 0.955 = 1.910 M

(b) Number of atoms = number of moles X N_A X atomicity

                          = 0.1 X 6.02 X 10²³ X 3

                          = 1.806 X 10²³ atoms

(a) Number of milliequivalents of HCl = 20 X 0.050 X 1 = 1

Number of milliequivalents of Ba(OH)2 = 2 X 30 X 0.10 = 6

[OH^-] of final solution $=\frac{\mathrm{milliequivalents} \mathrm{of} \mathrm{Ba}(\mathrm{OH}{)}_2 - \mathrm{milliequivalents} \mathrm{of} \mathrm{HCl}}{\mathrm{total} \mathrm{volume}}$

                                 $=\frac{6-1}{50}=0.1 \mathrm{M}$

(c) Key Idea (i) ${\mathrm{H}}_2 +\hspace{0.17em}\frac{1}{2}{\mathrm{O}}_2 \stackrel{ }{\to } {\mathrm{H}}_2\mathrm{O}$

(ii) Amount of water produced is decided by limited reactant (i.e., the reactant which is used in small amount)

$$\begin{gathered} {\mathrm{H}}_2 + \hspace{0.17em}\frac{1}{2}{\mathrm{O}}_2 \stackrel{ }{\to } {\mathrm{H}}_2\mathrm{O} \\ 1\mathrm{mol} \frac{1}{2}\mathrm{mol} 1\mathrm{mol} \\ \frac{10}{2}\mathrm{mol} \frac{64}{32}\mathrm{mol} ? \\ =5\mathrm{mol} =2\mathrm{mol} \\ \because \frac{1}{2}\mathrm{mol} {\mathrm{O}}_2 \mathrm{gives} = 1\mathrm{mol} {\mathrm{H}}_2\mathrm{O} \\ \therefore 2 \mathrm{mol} {\mathrm{O}}_2 \mathrm{will} \mathrm{give} = 1\times 2\times 2 = 4\mathrm{mol}. \end{gathered}$$

6.023 x 10²³ molecules of water = 1mol =18g

.^.. Mass of one molecule of water = 18/6.023 x10²³ g

 ._.. d=m/V

.^.. V=m/d =18/(6.023x10²³x1) $\approx$3x10^-23 cm³

Key Idea: The reagent which is present in smaller quantity is called the limiting reagent and the moles of product depends on it and number of moles = weight/molecular weight

PbO        +        2HCl  $\to$   PbCl₂  +    H₂O

207.2 +16     2(35.5+1)    207.2+71

 =223.2            =73            =278.2

Mole of PbO = 6.5/223 = 0.029

Mole of HCl = 3.2/36.5 = 0.087

Here, 1 mole of PbO reacts with 2 moles of HCl, thus PbO is the limiting reagent.

._.. 223.2 g PbO gives PbCl₂ = 278.2g

.^.. 6.5g PbO will give PbCl₂

                              =(278.2/223.2) x 6.5g

                              =(278.2 x 6.5)/(223.2 x278.2) mol

                              = 0.029 mol

(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