Chemistry MCQs for NEET — Practice Questions with Answers

(a) Given, molarity of solution=2

Volume of solution = 250mL=250/1000=1/4L

Molar mass of HNO₃ = 1+14+3x16=63g mol^-1

Molarity = $\frac{weight of HN{O}_3}{Molecular mass of HN{O}_3\times volume of solution\left(L\right)}$

Weight of  HNO₃ = molarity x molecular massx volume(L) = 2x63x1/4g=31.5g

It is the weight of 100%  HNO₃

But the given acid is 70% HNO₃

Its weight = 31.5x100/70g=45g

(a) 1.00 molal aqueous solution = 1.0 mole in 1000g water

n_solute = 1; W_solvent =1000g

n_solvent = 1000/18 = 55.56

X_solute = $\frac{n_{solute}}{n_{solute}+n_{solvent}}$

X_solute = 1/1+55.56 = 0.0177 

Alternate Method

Molality(m) = 1000xn/NxM

where, n = number of moles of solute

         N = number of moles of solvent

        M = molar mass of solvent

Given, m=1

.^.. 1= 1000xn/Nx18 $\Rightarrow$n/N =18/1000

or n/n+N = 18/1018 =0.0177

(b) Given, number of molecules of urea= 6.02x10²⁰ 

Number of moles = 6.02x10²⁰ /N_A

=6.02x10²⁰ /6.02x10²³ =1x10^-3 mol

Volume of the solution = 100mL = 100/1000L =0.1L

Concentration of urea solution (in mol L^-1)

=1x10^-3 /0.1 mol L^-1

=1x10^-2 mol L^-1

=0.01 mol L^-1

(b) In 7g nitrogen, number of  molecules = $\frac{7.0}{28}$mol = 0.25 x N_A molecules

where, N_A = Avogadro number = 6.023 x 10²³

In 2g of H₂= 2/2 mol =1x N_A molecules

In 16 g of  NO₂= 16/46 mol = 0.348 x N_A molecules

In 16 g of O₂ = 16/32 mol = 0.5 x N_A molecules

Hence, maximum number of molecules are present in 2g of H₂.

(a) Volume oxygen in 1L of air = 21/100x1000 = 210mL

22400 mL volume at STP is occupied by oxygen = 1mole

Therefore, number of moles occupied by 210mL = 210/22400 = 0.0093mol

(c) C₃H₈ + 5O₂  $\to$3CO₂+ 4H₂O

22.4L   5x22.4L

For the combustion of 22.4L propane, oxygen required = 5 x 22.4L

For  the combustion of 1L of propane oxygen required = 5x22.4/22.4L

=5L

(d) Suppose the molecular weight of enzyme =  x

0.5% by weight means in 100g of enzyme weight of Se=0.5g

In xg of enzyme weight of Se = $\frac{.5}{100}\times x$

Hence, 78.4 = $\frac{0.5 \times x}{100}$

x = 15680 = 1.568x10⁴

0.6

10FeC₂O₄ + 6KMnO₄ + 24H₂SO₄ ----> 3K₂SO₄+ 6MnSO₄ +  5Fe₂(SO4)₃ + 24H₂O + 20CO₂

So we see that  6 moles of KMnO₄ is required to oxidize 10 moles of  FeC₂O₄  Then, 1 mole of FeC₂O₄ would be oxidized by = ?    6/10 = 0.6

(c) Moles of ${\mathrm{N}}_3^{-}$ ion = 4.2/42=0.1

Each nitrogen atom has 5 valence electrons. Therefore, total number of electrons of ${\mathrm{N}}_3^{-}$ ion=16

Total number of electrons in 0.1 mole or 

4.2g of ${\mathrm{N}}_3^{-}$ ion= 0.1x16xN_A 

=1.6N_A